%\documentclass{amsart} %\UseRawInputEncoding \documentclass[ALCO,Unicode,published]{cedram} \usepackage{bm} %\newtheorem{Theorem}[equation]{Theorem} %\newtheorem{theorem}[equation]{Theorem} %\newtheorem{Proposition}[equation]{Proposition} %\newtheorem{proposition}[equation]{Proposition} %\newtheorem{lemma}[equation]{Lemma} %\newtheorem{Lemma}[equation]{Lemma} %\newtheorem{definitions}[equation]{Definitions} %\newtheorem{Corollary}[equation]{Corollary} %\newtheorem{corollary}[equation]{Corollary} %\newtheorem{conjecture}[equation]{Conjecture} %\newtheorem{fact}[equation]{\hspace{-3pt}\rm} %\theoremstyle{definition} %\newtheorem{definition}[equation]{Definition} %\theoremstyle{remark} %\newtheorem{Remark}[equation]{Remark} %\newtheorem{Remarks}[equation]{Remarks} %\newtheorem{hypothesis}[equation]{Hypothesis} %\newtheorem{coro}[equation]{Corollary} %\newtheorem{prop}[equation]{Proposition} %\newtheorem{exam}[equation]{Example} %\newtheorem{defi}[equation]{Definition} %\newtheorem{rema}[equation]{Remark} %\newtheorem{Example}[equation]{Example} %\newtheorem{Examples}[equation]{Examples} %\newtheorem{notation}[equation]{Notation} %\newtheorem{problem}[equation]{Problem} %\newtheorem{open}[equation]{Question} % %\numberwithin{equation}{section} % \equalenv{Lemma}{lemm} %\def\Proof {\noindent{\em Proof.} } \DeclareMathOperator{\Aut}{Aut} \title[Automorphism groups of Steiner triple systems] {Automorphism groups of Steiner triple systems } \author{\firstname{Jean} \lastname{Doyen}} \address{Universit\'e Libre de Bruxelles\\ Bruxelles 1050\\ Belgium} \email{jdoyen@ulb.ac.be} \author{\firstname{William} \middlename{M.} \lastname{Kantor}} \address{U. of Oregon\\ Eugene\\ OR 97403 \ and \ Northeastern U.\\ Boston\\ MA 02115} \email{kantor@uoregon.edu} \thanks{The research of the second author was supported in part by a grant from the Simons Foundation.} \keywords{Steiner triple system, Automorphism group} \subjclass{05B07, 05B25, 51E10} \datereceived{2018-08-10} \daterevised{2022-03-25} \dateaccepted{2022-03-26} \datepublished{2022-09-08} \begin{document} \begin{abstract} % \label{corollary} If $G$ is a finite group then there is an integer $M_G$ such that$,$ for $u\ge M_G$ and $u\equiv 1$ or $3$ {\rm(mod~6),} there is a Steiner triple system $U$ on $u$ points for which ${\rm Aut}\hspace{.5pt} U \hspace{.5pt}\cong G.$ \ % If $V$ is a Steiner triple system then there is an integer $N_V$ such that$,$ for $u\ge N_V$ and $u\equiv 1$ or $3$ $($mod~$6),$ there is a Steiner triple system $U$ on $u$ points having $V$ as an ${\rm Aut}\hspace{.5pt} U$-invariant subsystem such that ${\rm Aut}\hspace{.5pt} U\cong{\rm Aut} V $ and ${\rm Aut}\hspace{.5pt} U $ induces ${\rm Aut} \hspace{.5pt}V$ on $V$. \end{abstract} \maketitle \section{Introduction} \label{Introduction} Mendelsohn \cite{Me} proved that any finite group $G$ is isomorphic to the automorphism group of some Steiner triple system. In his proof he modified the Steiner triple system of points and lines of a projective space $PG(n,2)$, producing a system having $2^{n+1}-1$ points for some $n$. This leads to the natural question: what restrictions are there on the number of points of a Steiner triple system $U$ such that ${\rm Aut}\hspace{.5pt} U\cong G$? In order to admit $G$ as a group of automorphisms, $U$ cannot be too small: \begin{theorem} \label{corollary} If $G$ is a finite group then there is an integer $M_G$ such that$,$ for $u\ge M_G$ and $u\equiv 1$ or $3$ {\rm(mod~6),} there is a Steiner triple system $U$ on $u$ points for which ${\rm Aut}\hspace{.5pt} U \cong G.$ \end{theorem} As with most theorems of this sort, the proof does not distinguish between cyclic groups and simple groups. It is known that $M_G=15$ when $G=1$ \cite{LR}. Our arguments cannot deal with such small Steiner triple systems. The preceding theorem is an~immediate consequence of \cite{Me} and the following more general result, which is the main theorem of this paper: \begin{theorem} \label{main} If $V_{\scriptscriptstyle\star} $ is a Steiner triple system then there is an integer $N_{V_{\star}}$\! such that$,$ for $u\ge N_{V_{\star}}\!$ and $u\equiv 1$ or $3$ {\rm(mod~6),} there is a Steiner triple system $U$ on $u$ points having ${V_{\scriptscriptstyle\star}}$ as an ${\rm Aut}\hspace{.5pt} U$-invariant subsystem such that ${\rm Aut}\hspace{.5pt} U\cong{\rm Aut} {V_{\scriptscriptstyle\star}} $ and ${\rm Aut}\hspace{.5pt} U $ induces ${\rm Aut} \hspace{.5pt}{V_{\scriptscriptstyle\star}}$ on ${V_{\scriptscriptstyle\star}}$. \end{theorem} Cameron \cite{Ca} considered a similar question. He proved that, if ${V}$ is a Steiner triple system of order $v$ (\ie having $v$ points), and if $u>6v^2$ with $u\equiv 1$ or $3$ $($mod~$6),$ then there is~a Steiner triple system $U$ of order $u$ in which ${V}$ can be embedded in such a way~that every automorphism of ${V}$ can be extended to $U$. His proof and ours use a familiar and wonderful construction of Moore \cite{Mo} from the not-so-distant past that combines three Steiner triple systems to produce a fourth. The proof of Theorem~\ref{main} first enlarges $V_{\scriptscriptstyle\star}$, without changing its automorphism group, as part of a process to obtain a Steiner triple system $U$ having a rich geometry of $PG(m,2)$ subsystems for various $m$ (cf.~Remark~\ref{rich}). This process involves Lemma~\ref{direct product} and Proposition~\ref{VxD} (using \cite{Ka}), and leads to our key tool: Proposition~\ref{maximal}. The latter makes it straightforward in Proposition~\ref{Aut A} to determine the automorphism group of the Steiner triple system $U$ we construct. The ugly bookkeeping parts of the proof (in Section~\ref{Properties of $X,$ $Y$ and $V$} and especially in Section~\ref{Existence of $y$ and $x$}) ensure that we obtain all large $u$. Remark~\ref{bounds} contains poor bounds for $M_G$ and~$N_{V_{\star}}$, while Remark~\ref{comparison} comments on a difference between the bookkeeping approaches in \cite{Ca} and here. Results such as Theorem~\ref{corollary} are usually based on the action of $G$ having many regular point-orbits. This is very much not the situation for Theorem~\ref{main}: for our $U$ the size of every point-orbit of ${\rm Aut}\hspace{.5pt} {U}$ is 1 or the size of a point-orbit of ${\rm Aut} {V_{\scriptscriptstyle\star}}$ on the original Steiner triple system ${V_{\scriptscriptstyle\star}}$. There is also a result in \cite{Ca} concerning a {\em partial Steiner triple system} $V$ (a set of points, together with some triples of points, such that any two points are in at most one triple), and a partial Steiner triple $\,U$ having $V$ as a {\em subsystem} (so the points of $V$ are among the points of $ U$, and the triples of $V$ are precisely those triples of $U$ that are contained in $V$). It is shown in \cite{Ca} that there is a function $g$ such that, if ${V}$ is a partial Steiner triple system of order $v,$ and if $u>g(v)$ with $u\equiv 1$ or $3$ $($mod~$6),$ then there is~a Steiner triple system $U$ of order $u$ of which ${V}$ is a subsystem such that every automorphism of ${V}$ can be extended to $U$. In Section~\ref{Proof PSTS} we will use Theorem~\ref{main} to prove the following stronger result (along with variations): \begin{theorem} \label{PSTS} If $V $ is a partial Steiner triple system then there is an integer $N'_{V }$ such that$,$ for $u\ge N'_{V }$ and $u\equiv 1$ or $3$ {\rm(mod~6),} there is a Steiner triple system $U$ on $u$ points having $ V $ as an ${\rm Aut}\hspace{.5pt} U$-invariant subsystem such that ${\rm Aut} \hspace{.5pt}U\cong{\rm Aut} V $ and ${\rm Aut} \hspace{.5pt}U $ induces ${\rm Aut} V$ on $V$. \end{theorem} %\newpage \section{Background} \subsection{Moore's $XYV$} \label{XYV} We will use a 125 year old construction due to Moore \cite[p.~276]{Mo}.\footnote{Moore used it to produce two nonisomorphic STSs of any admissible order $>13$. Unfortunately, his method for proving nonisomorphism \cite[pp.~279--281]{Mo} has a significant gap.} (This construction is in many sources, such as \cite[p.~235]{LW} and \cite{Ca}.) Let $X\subset Y$ and $V$ be three STSs (\ie Steiner triple systems), and label $Y-X$ in any way by the elements of a cyclic group $A$ of order $|Y|-|X|$. (We always use $|Y| $ to denote the order of an STS $\,Y$.) Then $U\coloneqq X\cup (V\times A)$ is (the set of points of) an STS, with triples \begin{enumerate}[label=(M\arabic*)] \item\label{M1} those of $X$, \item \label{M2} $(v,a_1), (v,a_2)$ and $\begin{cases} x&$\!\!$ \mbox{if $a_1,a_2,x$ is a triple in $Y$, $a_i\in Y-X$, $x\in X$} \\ (v,a_3)& $\!\!$\mbox{if $a_1,a_2,a_3$ is a triple in $Y$, $a_i\in Y-X$, } \end{cases}$ and \item \label{M3} $(v_1,a_1), (v_2,a_2 ), (v_3,a_3)$ if $v_1,v_2,v_3$ is a triple in $V$ and $a_1a_2a_3=1$. \end{enumerate} The fact that $A$ is cyclic, not just abelian, is used in several places, most significantly in Lemma~\ref{labeling} and Proposition~\ref{Aut A}. Clearly $|U|=|X|+|V|(|Y|-|X|)$. \subsection {Enlarging $Y$} \label {$2Y+1$ construction.} \label{2Y+1} The STSs $X$ and $Y$ in the preceding section have unknown structure. While this does not matter for $X$, we will use elementary constructions to enlarge $Y$ in order to give it significant geometric structure (Lemmas~\ref{PG(2,2)-pointed} and~\ref{direct product}). Given an STS $\, Y_0$ there is a standard construction for an STS $\, 2Y_0+1$ on $2|Y_0|+1$ points, labeled using $Y_0\dot\cup Y_0'\dot\cup *_0 $ for a ``distinguished'' new point $*_0$ and a bijection $y \mapsto y'$ sending $Y_0 \to Y ' _0$, and triples of the~form $$\mbox{ \ \ \ \ $abc$ in $Y_0$ \ \ \ \ $*_0aa'$ etc. \ \ \ \ $a'b'c$ etc.} $$ Here $|2Y_0+1|\equiv 3$~(mod~4) and $Y_0$ is a subsystem of $2Y_0+1$. If $Y_0$ is a hyperplane of a projective space $P=PG(n,2)$ then $P\cong 2Y_0+1$. Thus, if $Y_0$ is a projective space then so is $2Y_0+1$. \begin{definition} An STS is~{\em $PG(2,2)$-pointed} with respect to a point $p$ if any two triples containing $p$ generate a $PG(2,2)$ subsystem. %\medskip An STS with more than seven points is {\em $PG(3,2)$-$2$-pointed} with respect to two points if any four points including these two generate a $PG(k,2)$ subsystem for $k=2 $ or 3. An STS is \emph{$PG(3,2)$-paired} if any two points are in a $PG(3,2)$ subsystem. {\em This is the key geometric property} {\em needed in the proof of} Proposition~\ref{maximal}(i). \end{definition} \begin{Lemma} \label{PG(2,2)-pointed} If $Y_0$ is an STS with more than one point$,$ then \begin{itemize} \item[(i)] $2Y_0+1$ is $PG(2,2)$-pointed with respect to the distinguished point $*_0,$ \item[(ii)] $Y_1\coloneqq 2(2Y_0 + 1) + 1$ is $PG(3,2)$-$2$-pointed {$($with respect to some pair of its points$),$} and \item[(iii)] $Y_1$ is $PG(3,2)$-paired of order $ {4|Y_0 |+3}\equiv 7$~{\rm (mod~$8$).} \end{itemize} \end{Lemma} \begin{proof} (i) Two triples $*_0aa'$, $*_0bb'$ of $ 2Y_0 + 1 $ containing $*_0$ generate a $PG(2,2)$ subsystem with triples $$\mbox{ \ \ \ \ $abc$ \ \ \ \ \ \ $*_0aa'$, $*_0bb'$, $*_0cc'$ \ \ \ \ \ \ $a'b'c$, $a'bc'$, $ab'c'$. } $$ (ii) The pair $\{ *_0,*_1\}$ has the required property, where $*_1$ is the new point used to produce $ Y_1= {2 (2Y_0 +1)}+1 $ from $2Y_0 +1$. For, if $*_0,*_1,a,b$ are four points of $Y_1$ then $*_1,a $ and $*_1, b$ are in triples meeting $2Y_0 +1 $ at points $\bar a $ and $\bar b$, respectively. Then $*_0,\bar a,\bar b$ are in a $PG(2,2)$ subsystem $Z$ of $2Y_0+1$, and $2Z+1$ is a $PG(3,2)$ subsystem of $ Y_1 $ containing $*_0,*_1,a,b$. (iii) This is immediate by (ii). \end{proof} \medskip {\em Admissible integers} are those $\equiv1$ or 3 (mod 6); these are precisely the possible orders of STSs. We use the standard {\em direct product} \mbox{$A{\bm\times}B$} of STSs $A$ and $B$: the STS for which $A{\times} B$ is the set of points and whose triples have the form $\{(a,b_1),(a,b_2),(a,b_3 ) \}$, $\{ (a_1,b), (a_2,b), (a_3,b)\}$ or $\{(a_1,b_1),(a_2,b_2),(a_3,b_3)\}$ for $ a\in A, $ $b\in B$, and ordered triples $(a_1,a_2,a_3)$ from $A$ and $(b_1,b_2,b_3)$ from $B$. Clearly, if $a\in A$ and $b\in B$ then $a{\bm\times}B$ and $A{\bm\times}b$ are subsystems of $A{\bm\times}B$. \begin{Lemma} \label{direct product}~ \begin{itemize} \item [\rm(i)] If $A$ and $B$ are $PG(3,2)$-paired STSs then so is \mbox{$A{\bm\times}B$}. \item [\rm(ii)] An integer is the order of a $PG(3,2)$-paired STS if it is admissible$,$ at least $15$ and ${\equiv 7}$~{\rm(mod~8),} or if it is a product of integers each of which behaves that way. \end{itemize} \end{Lemma} \begin{proof} (i) Given two points $(a_1,b_1)$ and $(a_2,b_2)$ of \mbox{$A{\bm\times}B$} there are $PG(3,2)$ subsystems $A_0$ of $A$ containing $a_1,a_2$ and $B_0$ of $B$ containing $b_1,b_2$. If $a_1\ne a_2$ and $b_1\ne b_2$ there is an isomorphism $\phi\colon \!A_0\to B_0$ sending $a_i\mapsto b_i$ for $i=1,2$.~Then $\{(a ,a ^\phi ) \! \mid \!a\in A_0 \}\subset A_0{\bm\times} B_0$ is a $PG(3,2)$ subsystem of \mbox{$A{\bm\times}B$} containing the given~points. If $a_1= a_2$ then $a_1 {\bm\times} B_0$ is a $PG(3,2)$ subsystem containing $(a_1,b_1)$ and $(a_2,b_2)$. If $b_1= b_2$ then $A_0 {\bm\times} b_1$ is a $PG(3,2)$ subsystem containing $(a_1,b_1)$ and $(a_2,b_2)$. %\smallskip (ii) Any admissible integer $8n+7 \ge 15$ can be written $2(2(2n+1)+1)+1$ with $2n+1>1$ and admissible. By Lemma~\ref{PG(2,2)-pointed}(iii) there is a $PG(3,2)$-paired STS of order $8n+7$. Now use (i). \end{proof} % \smallskip\smallskip % \newpage \subsection{Enlarging ${V_{\scriptscriptstyle\star}}$} \label{Increasing $V$.} The STS $\, {V_{\scriptscriptstyle\star}}$ in Theorem~\ref{main} has unknown structure. As we did with $Y_0$ in Section~\ref{2Y+1}, we will enlarge $\, {V_{\scriptscriptstyle\star}}$ to STSs having significant geometric structure (Corollary~\ref{=3}). Since our arguments are based on finite geometry, we briefly describe how ``close'' each STS in \cite{Ka} is to a projective space. Let $V$ be a vector space over $F={\bf F}_{16}$ with basis $v_1,\dots,v_n.$ This produces a vector space and a projective space over ${\bf F}_2$. Let $F^*=\langle \theta\rangle$. Suitably modify the $PG(3,2)$ subspaces determined by the ${\bf F}_2$-spaces $Fv_i$ and $F(v_i+\theta v_j)$ for all $i,j,$ in order to obtain an STS $D$. Only a tiny portion of the underlying $PG({4n-1},2)$ is altered: every subspace of that projective space meeting $U\coloneqq \bigcup_{i,j} \big (Fv_i\cup {F(v_i+\theta v_j)} \big)$ at most once is a subsystem of $D$. As in \cite[Theorem 1.1 and Sec.~7(1c)]{Ka}, this provides the flexibility needed for the following \begin{Remark} \label{D}\em If $n\ge 6$ then there is an STS $\,D$ such that \begin{itemize} \item [(i)]${\rm Aut} D=1,$ \item [(ii)] $D$ has $16^n-1$ points$,$ and \item [(iii)] given points $a,b$ of $D$ there is a point $c$ such that each pair $\{a,c\}$ and $\{b,c\}$ is in some $PG(n-1,2)$~subsystem of $D$. \rm \end{itemize} \end{Remark} See \cite{Ka} for the modifications of the above $PG(3,2)$ subsystems needed to obtain (i). The subsystems in (iii) are crucial for our proof of Theorem~\ref{main}; we will obtain them less tediously than similar ones in \cite{Ka}. Let $c\notin \bigcup_{u\in U} ( \langle a,u \rangle _{{\bf F}_2} \cup \langle b,u \rangle _{{\bf F}_2} )$ and consider the pair $\{a, c \}$. For $2\le j|{V_{\scriptscriptstyle\star}}| \ge 2^{n-6} $.} Then \emph{the image of any map~from $PG(n-1,2)$ into ${V_{\scriptscriptstyle\star}}$ sending every collinear triple to a triple or a point must have size} 1. (Otherwise, since the map cannot be 1-1, restrict to a plane mapping onto a triple in order to obtain a contradiction: the preimages of the points of the triple would have to be pairwise disjoint, cover the plane, and be such that the line through two points of a preimage is contained in the preimage.) \begin{proposition} \label{VxD} Let $D$ be as in the preceding Remark, and let $d\in D $. \begin{itemize} \item[(i)] Every point of ${V_{\scriptscriptstyle\star}}{\bm\times} D$ is in a $PG(n-1,2)$ subsystem of ${V_{\scriptscriptstyle\star}}{\bm\times} D,$ \item[(ii)] ${V_{\scriptscriptstyle\star}}{\bm\times} d $ is isomorphic to ${V_{\scriptscriptstyle\star}},$ \item[(iii)] ${V_{\scriptscriptstyle\star}}{\bm\times} d $ is an ${\rm Aut}({V_{\scriptscriptstyle\star}}{\bm\times} D)$-invariant subsystem of ${V_{\scriptscriptstyle\star}}{\bm\times} D$ on which ${\rm Aut}({V_{\scriptscriptstyle\star}}{\bm\times}D)$ induces $ {\rm Aut} ({V_{\scriptscriptstyle\star}}{\bm\times} d),$ and $ {\rm Aut} ({V_{\scriptscriptstyle\star}}{\bm\times} D) \cong {\rm Aut} ({V_{\scriptscriptstyle\star}}{\bm\times} d) \cong {\rm Aut} \hspace{.5pt} {V_{\scriptscriptstyle\star}} ,$ and \item[(iv)] $|{V_{\scriptscriptstyle\star}}{\bm\times} D|\equiv \pm3$ {\rm (mod~12)} and $16^5<|{V_{\scriptscriptstyle\star}}{\bm\times} D|<2^{ 24} | {V_{\scriptscriptstyle\star}}|^5$. \end{itemize} \end{proposition} \begin{proof} For (i), if $v\in {V_{\scriptscriptstyle\star}}$ then $v{\bm\times}D $ is isomorphic to $D$, so use Remark~\ref{D}(iii). Statement (ii) is obvious. For (iii) we need to determine ${\rm Aut}({V_{\scriptscriptstyle\star}}{\bm\times}D)$. Every $PG(n-1,2)$ subsystem in ${V_{\scriptscriptstyle\star}}{\bm\times}D$ is a set of ordered pairs that projects onto subsystems of ${V_{\scriptscriptstyle\star}}$ and $D$, and hence induces a map from $PG(n-1,2)$ to $V_{\scriptscriptstyle\star}$ sending every collinear triple to a point or triple. As noted above, since $2^n-1> |{V_{\scriptscriptstyle\star}}| $ the image of that map is a point of ${V_{\scriptscriptstyle\star}}$. Thus, {\em every $PG(n-1,2)$ subsystem in ${V_{\scriptscriptstyle\star}}{\bm\times} D$ lies in some subsystem $v{\bm\times} D,$ $ v\in {V_{\scriptscriptstyle\star}}$}. Consider the graph whose vertices are the $PG(n-1,2)$ subsystems of ${V_{\scriptscriptstyle\star}}{\bm\times} D$, with two such subsystems joined if and only if they meet. We have just seen that every such subsystem is contained in some subsystem $v{\bm\times} D$, $v\in V$. Since no two subsystems $v{\bm\times} D$ meet, every connected component $C$ of our graph lies in some subsystem $v{\bm\times} D;$ by Remark~\ref{D}(iii), $C$ is the set of $PG(n-1,2)$ subsystems of $v{\bm\times} D$ and generates $v{\bm\times} D$. Since ${{\rm Aut}({V_{\scriptscriptstyle\star}}{\bm\times}D) }$ permutes the connected components of our graph it also permutes the subsystems $v{\bm\times} D$. Let $h\in {\rm Aut}({V_{\scriptscriptstyle\star}}{\bm\times} D)$. If $v\in V_{\scriptscriptstyle\star}$ then $(v{\bm\times}D)^h= v^{h'}{\bm\times}D $ for some $v^{h'}\in V_{\scriptscriptstyle\star}$, where $h'\colon V_{\scriptscriptstyle\star}\to V_{\scriptscriptstyle\star}$ is bijective. We claim that the map $h'\colon \!{V_{\scriptscriptstyle\star}}\to {V_{\scriptscriptstyle\star}}$ {\em is in} ${\rm Aut}V_{\scriptscriptstyle\star}$.~~For, if $v_1,v_2,v_3$ is a triple of $V_{\scriptscriptstyle\star}$ and $d\in D$ then $(v_1,d),(v_2,d),(v_3,d)$ is a triple of ${V_{\scriptscriptstyle\star}}{\bm\times} D$. Then so is $(v_1,d)^h,(v_2,d)^h,(v_3,d)^h$; this is $(v_1^{h'},d_1),(v_2^{h'},d_2),(v_3^{h'},d_3)$ with $d_i\in D$, so $v_1^{h'},v_2^{h'},$ $ v_3^{h'}$ is a triple of $V_{\scriptscriptstyle\star}$, as claimed. Now $h' \in {\rm Aut}\hspace{.5pt} V_{\scriptscriptstyle\star}$ induces $h^\bullet \in {{\rm Aut}({V_{\scriptscriptstyle\star}}{\bm\times}d) }$ sending $(v,d)\mapsto (v^{h'},d)$ for ${v\in {V_{\scriptscriptstyle\star}}},$ ${d\in D}$. We thus have two automorphisms $h$ and $h^\bullet$ of $ {V_{\scriptscriptstyle\star}}{\bm\times}d $ sending each $v{\bm\times} D $ to $v^{h'}{\bm\times} D $. Then $ h^\bullet h^{-1}$ sends each subsystem $ v{\bm\times} D $ to itself, induces an automorphism of each such subsystem, and hence is 1 by (ii) and Remark~\ref{D}(i). Thus, $ h=h^\bullet $ sends each subsystem $ {V_{\scriptscriptstyle\star}}{\bm\times} d $ to itself, and hence so does ${\rm Aut}({V_{\scriptscriptstyle\star}}{\bm\times}D)$. This proves that ${\rm Aut}({V_{\scriptscriptstyle\star}}{\bm\times}D)= ( {\rm Aut}{V_{\scriptscriptstyle\star}})\times 1_D$. (iv) $|V_{\scriptscriptstyle\star}{\bm\times}D| = (16^n-1)|{V_{\scriptscriptstyle\star}}|\equiv (16-1) |{V_{\scriptscriptstyle\star}}|\equiv 3|{V_{\scriptscriptstyle\star}}|\equiv \pm3$~{\rm (mod~12)} and $16^5<{(16^n-1)}|{V_{\scriptscriptstyle\star}}|< (2^n)^4| {V_{\scriptscriptstyle\star}} | \le (2^{ 6} | {V_{\scriptscriptstyle\star}} |)^4 | {V_{\scriptscriptstyle\star}} |$ since $| {V_{\scriptscriptstyle\star}} |\ge 2^{n-6}$. \end{proof} %\bigskip \begin{corollary} \label{=3} For sufficiently large integers $n$ and $m$ there are STSs $V_{\scriptscriptstyle\star}{\bm\times}D$ and $V_{\scriptscriptstyle\star}{\bm\times}D{\bm\times}D'$ such that one of them$,$ $V,$ has the following properties$:$ \begin{itemize} \item[(i)] $|V|\equiv 3 $ {\rm (mod~12),} $|V|>16^5,$ and either $|V|=|V_{\scriptscriptstyle\star}||D|=|V_{\scriptscriptstyle\star}| (16^n-1) $ or $|V|=|V_{\scriptscriptstyle\star}||D||D'|=|V_{\scriptscriptstyle\star}| (16^n-1)(16^m-1) ,$ \item[(ii)] Every point of $V$ is in a $PG(3,2)$ subsystem of $V,$ and \item[(iii)] $ V$ has an ${\rm Aut}\hspace{.5pt} V$-invariant subsystem ${V_{\scriptscriptstyle\star}}' \cong{V_{\scriptscriptstyle\star}} $ on which ${\rm Aut}\hspace{.5pt} V$ induces~${\rm Aut}\hspace{.5pt} {V_{\scriptscriptstyle\star}}' ,$ and $ {\rm Aut} V\cong {\rm Aut} {V_{\scriptscriptstyle\star}}' \cong {\rm Aut} {V_{\scriptscriptstyle\star}}.$ \end{itemize} \end{corollary} \begin{proof} Apply the proposition to ${V_{\scriptscriptstyle\star}}{\bm\times} D$ in place of ${V_{\scriptscriptstyle\star}}$, using in place of $D$ an STS $D'$ of order $16^m-1$ where $2^m-1>|{V_{\scriptscriptstyle\star}}{\bm\times} D| \ge 2^{m-6}$. Since $3 |V_{\scriptscriptstyle\star}|\equiv\pm3$ (mod 12) either $ |V_{\scriptscriptstyle\star}|(16^n-1)\equiv3$ (mod 12) or $ |V_{\scriptscriptstyle\star}|(16^n-1)(16^m-1) \equiv3$ (mod 12). \end{proof} %\newpage \section{Proof of Theorem~\ref{main}} \label {Proof} The proof proceeds in three stages: construction of an STS $U$ (Section~\ref{Properties of $X,$ $Y$ and $V$}), determining that we have obtained all sufficiently large admissible integers as the order of some such $U$ (Section~\ref{Existence of $y$ and $x$}), and using geometry to determine {\rm Aut\hspace{.5pt}}$U$ (Sections~\ref{Finding critical subsystems} and \ref{The occurrence of $A$.}). \subsection {Preliminaries and notation} \label{Preliminaries} We will describe STSs~$X,$ $Y$ and $V$ that will be used to construct our STS~$U$ via Section~\ref{XYV}. \subsubsection {Properties of $X,$ $Y$ and $V$} \label {Properties of $X,$ $Y$ and $V$} We begin with notation and properties of these STSs. We note that properties~\ref{p1}(b) and~\ref{p3}(d) will be essential (in Proposition~\ref{maximal}) for studying subsystems of $U$ isomorphic to $V$ or $Y$. Let $V_*$ be as in Theorem~\ref{main}. Property~\ref{p1} concerns an STS $\,V$ that will replace $V_*$ in our arguments and has a rich geometric structure. % \smallskip % \newpage \begin{enumerate}[label=(P\arabic*)] \item \label{p1} {\em$V$ and $v$.} \begin{itemize} \item [(a)] Use Corollary~\ref{=3} to obtain an STS $\,V$ on $v$ points, where \qquad\qquad $v\equiv 3 $ {\rm (mod~12)} and $ v>16^5$, having (a copy of) ${V_{\scriptscriptstyle\star}}$ as an ${\rm Aut} V $-invariant subsystem on which ${\rm Aut} \hspace{.5pt} V $ induces ${\rm Aut}\hspace{.5pt} {V_{\scriptscriptstyle\star}}$ and such that ${\rm Aut} \hspace{.5pt} V \cong {\rm Aut} {V_{\scriptscriptstyle\star}}.$ (Here $V $ is not uniquely determined: it depends on choices for $D,$ $n$ and possibly $D'$ and $m$ that are made just once in the proof of Corollary~\ref{=3}.) \item [(b)] Every point of $V$ is in a $PG(3,2)$ subsystem of $V$ (Corollary~\ref{=3}(ii)). \end{itemize} % \medskip \end{enumerate} In~\ref{p2} and ~\ref{p4} we will introduce further admissible integers $y_1$, $y_2$, $y$, $x$ and $u$; Lemma~\ref{Existence2} concerns the existence of integers satisfying the conditions stated in ~\ref{p1}, ~\ref{p2} and ~\ref{p4}. % \smallskip % \newpage \begin{enumerate}[label=(P\arabic*)] \addtocounter{enumi}{1} \item \label{p2} {\em$\delta, $ $x$, $y_1,$ $ y_2$ and $y$.} \begin{itemize} \item [(a)] Let $\delta=\pm1$. (The admissible integer $u$ in Theorem~\ref{main} will later be related to~$\delta$ via the requirement $u \equiv\delta $~(mod~4).) We will use an admissible integer $y \equiv\delta $~(mod~4): \ \ \ let $y_1\equiv 15$~(mod~24), which is admissible and $ \equiv 7$~(mod~8); \ \ \ if $\delta=-1$ let $y\coloneqq y_1$; and \ \ \ if $\delta= 1$ let $y\coloneqq y_1y_2$ where $y_2\equiv15$~(mod~24). \item [(b)] Let $x$ be admissible. \item [(c)] Assume that $y_1\ge \sqrt y\ge 8x+7$. \item [(d)] Assume that $y>x+ 6 v$. \end{itemize} \end{enumerate} In~\ref{p3} and~\ref{p4} we provide a recipe that uses the integers in~\ref{p2} to obtain auxiliary STSs together with an STS $\,U$ that behaves as required in Theorem~\ref{main}. \begin{enumerate} [label=(P\arabic*)] \addtocounter{enumi}{2} %\begin{itemize} \item \label{p3} {\em$X,$ $Y_1,$ $Y_2$ and $Y$.} \begin{itemize} \item [(a)] Write $y_1=4y_0+3$, so that $y_0$ is admissible and $y_0\ge2x+1$ (by~\ref{p2}(a) and~\ref{p2}(c)). Then \cite{DW} provides an STS $\,Y_0$ of order $y_0$ containing a subsystem $X$ of order $x$. \item [(b)] Let $Y_1\coloneqq 2(2Y_0+1)+1$, so $ |Y_1|=2(2|Y_0|+1)+1=y_1$ (by \ref{p3}(a)) and $Y_1$ is $PG(3 ,2)$-paired (by Lemma~\ref{PG(2,2)-pointed}(iii)). If $\delta=1$ let $Y_2$ be a $PG(3 ,2)$-paired STS of order $y_2$ (Lemma \ref{direct product}(ii)). \item [(c)] If $\delta=-1$ let $Y\coloneqq Y_1$, of order $y$. If $\delta=1$ let $Y\coloneqq Y_1{\bm\times} Y_2$, of order $y$. \item [(d)] Any two points of $ Y$ are in a $PG(3,2)$ subsystem of $ Y$ (Lemma~\ref{direct product}). \end{itemize} % \end{itemize} % \newpage %\medskip % \begin{itemize} \item \label{p4} {\em$U,$ $A$ and $u$.} \noindent Let $ A\coloneqq Y-X$ be as in Section~\ref{XYV}, so the STS $U\coloneqq X\cup (V\times A)$ has order~$u\coloneqq x+v(y-x)$. In Lemma~\ref{Existence2} we will see that all sufficiently large admissible integers arise here as $u$ for the choice of $v$ in ~\ref{p1} and for suitable $x,y,\delta$ in ~\ref{p2}. As noted in \cite[p.~469]{Ca}, each $g\in {\rm Aut} V$ acts as an automorphism of $U$ via% \vspace{-2pt} $$ \mbox{$g=1$ on $X$ and $(p,q)^g=(p^g,q)$ for $p\in V,$ $ q\in Y-X$.}\vspace{-2pt} $$ This produces a subgroup of ${\rm Aut}\hspace{.5pt} U$ isomorphic to ${\rm Aut} \hspace{.5pt} V$ and inducing ${\rm Aut} \hspace{.5pt} (V\times 1)$ on the subsystem $V\times 1$ of $U$. % \smallskip \item \label{p5} The cyclic group $ A$ has even order; let $-1$ denote its involution. For $a\in A$ let $-a\coloneqq (-1)a$. Let $ A_6\coloneqq \{a\in A\mid a^6=1\}$. \smallskip \item \label{p6} {\em Labeling $ Y-X$.} We assume that $Y-X$ behaves as in Lemma~\ref{labeling} below (which depends only on Section~\ref{XYV} and the fact that $|Y-X|$ is not tiny). \end{enumerate} \begin{Remark} \label{rich} Section~\ref{Introduction} mentions ``{\em\dots a Steiner triple system $U$ having a rich geometry of $PG(m,2)$ subsystems\dots}''. This refers to the geometry inherited by $U$ from \ref{p1}(b) and \ref{p3}(d), which involves far more structure than the fact that $U$ is generated by its $PG(3,2)$ subsystems. \end{Remark} \subsubsection {Existence of $y$ and $x$} \label {Existence of $y$ and $x$} We first rephrase and slightly strengthen the numerical requirements in ~\ref{p1}, ~\ref{p2} and ~\ref{p4}: % \newpage \begin{Lemma} \label{Existence1} Assume that $u,v,\delta , y_1,y_2$ and $y$ are integers that behave as follows$:$ \begin{enumerate} \item [\rm(i)] $u$ is admissible with $u \equiv \delta {~\rm(mod~4)}$ for $\delta = \pm1,$ and $v\equiv 3$ {\rm(mod 12)}$,$ \item [\rm(ii)] $u> 800^2v^7$ and $v>16^5 , $ \item [\rm(iii)] $v-1$ is a factor of $u-y , $ \item [\rm(iv)] $y=y_1y_2 \equiv \delta {~\rm(mod~4)}$ with $y_1 \equiv 15${~\rm(mod~24)}$ ,$ where \noindent if $\delta=-1$ then $y_2=1,$ while \noindent if $\delta=1 $ then $y_ 2 \equiv 15${~\rm(mod~24)} and $y_1\ge y_2 $ $($so $y_1\ge \sqrt y\hspace{1pt}) ,$ and \item [\rm(v)] ${u/v}< y <{u/v} +\frac{1}{8}\{(v-1)/v\}\sqrt {{u/v}}\, -1 0$ and $u- y>0$, so $0u/v > 12 v $ by (v) and (ii); and this proves~\ref{p2}(d). \smallskip\smallskip \ref{p4}: The relation $u=x+v(y-x)$ is just the present definition of $x$. \end{proof} %\newpage \medskip %\smallskip {\em\noindent Remark.} Condition (v) places $y$ in an interval of length roughly $ \frac{1}{8} \sqrt {{u/v}} $, which is fairly large by (ii). We still need to verify the relatively obvious fact that this is large enough to make it possible to satisfy the remaining inequalities and congruences in the lemma. \begin{Lemma} \label{Existence2} Given admissible integers $v$ and $u$ such that {\rm$v\equiv 3$~(mod~12),} $v>16^5$ and $u\!>\! 800^2v^7 ,$ there are integers $x,y_1,y_2$ and $y$ behaving as stated in ~\ref{p1},~\ref{p2} and~\ref{p4}. \end{Lemma} \begin{proof} We will use (i)--(v) in the preceding lemma. Let $u\equiv \delta$~(mod~4) with $\delta=\pm1$; the remaining requirements in (i) and (ii) are among the present hypotheses. Since $v\equiv 3$~(mod~12) we can write $v = 3 +12e+24m$ with $e\in\{0,1\}$. When $\delta = -1 $ let $y_2\coloneqq 1$. When $\delta = 1 $ let $y_2\coloneqq v+(6-e)(v-1)$, so $y_2 \equiv 1$~(mod~$v-1$) and $y_2 \equiv 15$~(mod~$24$). Clearly $y_2<7v$. We next define $y_1$. Let $024v(v-1)7v>24v(v-1)y_2$ implies that $y_1>24(v-1)>7v>y_2$. %\smallskip (v): For the first part of (v), ${y=y_1y_2}> {\big (24 u /\{24 v(v-1) y_2 \}\big)}(v-1) y_2 =u/v$. Next, $u' 6\cdot 16^5$. Then there are distinct points $y_1,\dots,y_9$ of $Y-X$, and $x _0\in X$, such that the following are triples of $Y$: $$y_1,y_2,y_3 \qquad y_3,y_4,y_5 \qquad x _0,y_6,y_7 \qquad x _0,y_8,y_9.$$ Let $c$ be a generator of $A$. Label the $y_i$ using $A_6\cup \{c,-c,c^2\}$: \begin{equation*} \begin{array}{lllllll} &y_1=1 &y_2=-1& y_3=c& y_4=-c&y_5=c^2 ; \rm \ \ and \ also \vspace{4pt} \\ & y_6=\omega& y_7=-\omega&y_8=\omega^2&y_9=- \omega^2& \mbox{\rm if some $ \omega \in A $ has order 3.}& \end{array} \end{equation*} (The remaining points of $Y-X$ are labeled by the remaining elements of $A$ in an arbitrary manner. Note that the points $y_6,y_7,y_8$ and $y_9$ are needed only if $|A|$ is a multiple of 3.) Thus, we have the following triples in $Y$: $$ % \begin{equation} %\label{triples!} 1,-1,c \qquad c,-c,c^2 \qquad x _0,\omega, -\omega \qquad x _0,\omega^2, -\omega^2 %\end{equation} $$ (where the last two are omitted if $|A|$ is not a multiple of 3). Now consider an automorphism $y\mapsto k y^\alpha$ of $Y-X$, where $k\in A_6,$ $\alpha\in {\rm Aut} A $. This sends the triple $1,-1,c$ to $k,-k,kc^\alpha. $ If $k\in A_6-\{ \pm1\}$ then this triple is $\omega^i,- \omega^i,k c^\alpha$ for $i=1$ or 2; but the triple in $Y$ containing $\omega^i$ and $- \omega^i$ is not contained in $Y-X$. Thus, $k=\pm1$, and $1,-1,c$ is sent to $1,-1, k c^\alpha$, so $k c^\alpha=c$. If $k=-1$ then $c^\alpha=-c. $ The triple $c,-c, c^2$ is sent to $ -c^\alpha, -(-c)^\alpha, -(c^2)^\alpha$, which is $c, -c, -(c^\alpha)^{2} $. Since $-(c^\alpha)^{2}=-(-c)^{2}\ne c^{2}$, this is impossible. Then $k=1$, so $\alpha=1 $ since $c=c^\alpha$ generates $A$. \end{proof} \subsection {Location of $PG(2,2)$ subsystems} \label {Location of $PG(2,2)s$.} We need structural properties of the STS $\,U$ defined in ~\ref{p4}. In this section we will not need any of the assumptions in Section~\ref{Preliminaries}: only the definitions in ~\ref{p4} and the notation in ~\ref{p5} are involved. For $v\in V$ let $$ Y_v\coloneqq X\cup (v\times A) .$$ By~\ref{M2} this is a subsystem of $U$ isomorphic to $Y$ (via the isomorphism $x\mapsto x,$ $ y\mapsto (v,y)$ for $x\in X,$ $ y\in A=Y-X$). There are $PG(2,2)$ subsystems contained in $V\times 1$, and ones contained in $ Y_v \cong Y$ for $v\in V$. Another~possible type of $PG(2,2) $ subsystem uses a triple $v_1,v_2,v_3$ in~$V$, $x\in X$, and elements $a_i\in A$:% \vspace{-1pt} \begin{equation} \label{oddball} \begin{array}{lll} \mbox{points: $x,$ $ (v_i,a_i) ,$ $ (v_i,- a_i) $ for $i=1,2,3 , $} \\\mbox{\hspace{34.5pt}for $x\in X,$ $a_i\in A,$ $a_1a_2a_3=1$ and triples $a_i ,-a_i,x$ in $ Y$} \vspace{2pt}\\ \mbox{triples: $ (v_i, a_i) , (v_i, -a_i) , x$ for $i=1,2,3$, and} \vspace{2pt}\\ \hspace{34.5pt} \mbox{$ (v_1, \epsilon _1 a_1) , (v_2, \epsilon_2a_2) , (v_3, \epsilon_3a_3)$ whenever $\epsilon_i=\pm1$ and $\epsilon_1\epsilon_2\epsilon_3=1$.} \end{array} \end{equation} %\Remark \begin{Remark} \label{oddball determined}\em Two points $(v_1,a_1), (v_2,a_2)$ with $v_1\ne v_2$ lie in at most one subsystem \eqref{oddball}. \rm %\rm For, these points determine the triple $v_1,v_2,v_3$ and then all $(v_i,\pm a_i)$. \end{Remark} \begin{definition} \label{V_{}} Let $S$ be a subsystem of $ V$ and $f \colon \!S\to A_6$ (cf. ~\ref{p5}) a function such that $f({v_1})f({v_2})f({v_3})=1$ whenever $v_1,v_2,v_3$ is a triple in $S$. Then $$V_{S,f}\coloneqq \{ (s,f(s)) \mid {s \in S} \} $$ is a subsystem of $U$, and $V_{S,1}=S\times 1 \cong V_{S,f}$ via $(s,1)\mapsto (s,f(s))$, using ~\ref{M3}: these subsystems are just variations on the subsystem $V \times 1$. \end{definition} The subsystems $ Y_v$ and $V_{S,f}$ are basic tools in our proof of Theorem~\ref{main}, and \begin{equation} \label{int} \mbox{$| Y_v\cap V_{S, f}|\le 1$ for all $v,$ $S$ and $f$.} \end{equation} \begin{Lemma} \label{PG(2,2)s} Every $PG(2,2)$ subsystem of $U$ either is of type \eqref{oddball}, lies in some $ Y_v,$ or has the form $ V_{S, f} $ for a $PG(2,2)$ subsystem $S$ of $V$. \end{Lemma} \begin{proof} If a $PG(2,2)$ subsystem $Z$ has the form $\{(v_i,y_i)\mid 1\le i\le 7\}$ with distinct $v_i$, then the $v_i$ form an STS $ S$ by ~\ref{M3}; we may assume that the triples in $Z$ are $ \vspace{1pt} \hspace{21pt} \begin{array}{llll} (v_1,a_1) , (v_2,a_2) , (v_3,a_3) && \mbox{so\, $a_1a_2a_3=1$} \\ (v_1,a_1) , (v_4,a_4) ,(v_5,a_5) & & \mbox{so\, $a_1a_4a_5=1$} \hspace{130pt} \\ (v_1,a_1) , (v_6,a_6) ,(v_7,a_7) && \mbox{so\, $a_1a_6a_7=1$} \\ (v_3,a_3) , (v_5,a_5) ,(v_7,a_7) && \mbox{so\, $a_3a_5a_7=1$} \\ (v_3,a_3) , (v_4,a_4) , (v_6,a_6) && \mbox{so\, $a_3a_4a_6=1$} \\ (v_2,a_2) , (v_4,a_4) ,(v_7,a_7) && \mbox{so\, $a_2a_4a_7=1$} \\ (v_2,a_2) , (v_5,a_5) ,(v_6,a_6) && \mbox{so\, $a_2a_5a_6=1$} \end{array} $ \vspace{1pt} \vspace{1pt} \noindent with $a_i\in A$. Multiplying these equations, and also just the first three of them, we find that \vspace{2pt} $(\prod_ia_i)^3=1$ and $a_1^3 a_2a_3a_4 a_5a_6a_7=1$. It follows that $\prod_ia_i=\omega $ with $\omega^3=1$ and $a_1^2=\omega^2$, so every $a_i ^6=1$. Then $Z=V_{S,f}$ with $f(v_i)\coloneqq a_i\in A_6$. If $Z$ contains a triple $(v,a_1), (v,a_2), (v,a_3)$ but does not lie in $ Y_v$, then it also contains a point $(v_2,b_2)$ with $v_2\ne v$. By ~\ref{M3}, if $v,v_2,v_3$ is a triple of $V$ then there are triples $(v_2,b_2),(v,a_1),(v_3,b_3)$ and $(v_2,b_2),(v,a_2),(v_3,c_3)$ and $(v_2,b_2),(v,a_3),(v_3,d_3)$, and hence another triple $(v,a_1),(v_3,c_3),(v_3,d_3)$, which contradicts~\ref{M2}. Assume that $Z$ is not in any $ Y_v$. If $Z$ has a triple $T\subseteq X$, then some point $(v,a)$ is in $Z$, the triples joining $(v,a)$ to the points of $T$ all lie in both $Z$ and $Y_v$ by~\ref{M2}, and then $Z\subseteq Y_v$. The only other possibility is that $Z$ is determined by three triples through some $x\in X$, and hence contains triples $\hspace{21pt} \begin{array}{llll} (v_1,a_1) , (v_2,a_2) , (v_3,a_3) && \mbox{so $a_1a_2a_3=1$} \\ x, (v_1,a_1) , (v_1,b_1) && \mbox{so\, $a_1, b_1,x$\, is a triple in $ Y$} \\ x, (v_2,a_2) , (v_2,b_2) && \mbox{so\, $a_2, b_2,x$\, is a triple in $ Y$} \\ x, (v_3,a_3) , (v_3,b_3) && \mbox{so\, $a_3, b_3,x$\, is a triple in $ Y$} \\ (v_1,a_1) , (v_2,b_2) , (v_3,b_3) && \mbox{so\, $a_1b_2b_3=1$} \\ (v_1,b_1) , (v_2,b_2) , (v_3,a_3) && \mbox{so\, $b_1b_2a_3=1$} \\ (v_1,b_1) , (v_2,a_2) , (v_3,b_3) && \mbox{so\, $b_1a_2b_3=1$.} \end{array} \vspace{1pt} \vspace{1pt} $ \noindent The last three equations imply that $a_1a_2a_3b_1^2b_2^2b_3^2=1$, so $1=(b_1b_2b_3)^2=(b_ia_i^{-1})^2$ and hence $b_i=\pm a_i$ for each $i$ (cf. ~\ref{p5}). Since $ x, (v_i,a_i) , (v_i,b_i) $ is a triple we have $b_i=-a_i$ for each $i$, so we are in \eqref{oddball}. \end{proof} \medskip For Moore \cite[Sec.~10]{Mo}, the types of $PG(2,2)$ subsystems of $\,U$ were isomorphism invariants of his STS construction.\footnote{See Footnote 1.} He did not go into the detail involved in \eqref{oddball} or a function ${f\colon\! S\to A_6}$.% %\newpage \subsection{Finding critical subsystems} \label{Finding critical subsystems} %\medskip The following key result is based on \ref{p1}(b) and \ref{p3}(d) together with (\ref{int}). \begin{proposition} \label{maximal} Let $W$ be a subsystem of $U$. \begin{enumerate} \item[(i)] If $W\cong Y $ then $W$ has the form $ Y_v$ for some $v\in V$. \item[(ii)] If $W\cong V ,$ if $W$ meets every subsystem in {\rm (i)} in at most one point$,$ and if $W$ is disjoint from the intersection of the subsystems in {\rm (i),} then $W$ has the form $V_{V, f}$ for some ${f\colon\! V\to A_6}$. \end{enumerate} \end{proposition} \begin{proof} (i) Clearly $|W|=|Y|>|X|$. Let $(v_1,a_1)\in W-X$. We claim that $W \subseteq Y_{v_1}$. Assume that $W\not\subseteq Y_{v_1}$. Let $(v_2,a_2)\in W$ with $v_1\ne v_2$. Since $W\cong Y $, by~\ref{p3}(d) there is a $PG(3,2)$ subsystem containing $(v_1,a_1) $ and $ (v_2,a_2)$. That subsystem has three $PG(2,2)$ subsystems containing $(v_1,a_1)$ and $ (v_2,a_2)$, each of type $ V_{S, f} $ or as in \eqref{oddball}, by Lemma~\ref{PG(2,2)s}; and at least one has type $ V_{S, f} $ by Remark~\ref{oddball determined}. In particular, $a_1\in A_6$. Thus, $W\subseteq X\cup {(V \times A_6)}$.~Then $| Y|=|W|\le |X|+6|V|$, which contradicts~\ref{p2}(d). \ %\smallskip (ii) The stated intersection is $X$, so $W\subseteq V\times A$. For $v\in V$, $|W\cap Y_v|\le1$ implies that $v$ occurs in at most one pair $(v,a) \in W $. Since $|W|=|V|$, it follows that $W=\{(v,f(v)) \mid v\in V\} $ for some $f\colon \!V\to A$. Since $W\cong V $, by~\ref{p1}(b) every point of $W$ is in a $PG(2,2)$ subsystem, which by Lemma~\ref{PG(2,2)s} has the form $ V_{S, f'} $ with $|S|=7$ and $f '\colon \! S\to A_6$ (since $|W\cap Y_v|\le1$ for each $v$). Then $W\subseteq V\times A_6$, $f $ maps to $ A_6$, and by ~\ref{M3} $f$ must behave as in Definition~\ref{V_{}}. \end{proof} % \newpage \subsection{${\rm Aut}\hspace{.5pt} U$ and ${\rm Aut} A$} \label{The occurrence of $A$.} Theorem~\ref{main} concerns ${\rm Aut} \hspace{.5pt} U$: \begin{proposition} \label{Aut A} ${\rm Aut}\hspace{.5pt} U\cong {\rm Aut}\hspace{.5pt} V,$ and ${\rm Aut} \hspace{.5pt} U$ leaves $V\times 1$ invariant$, $ inducing ${\rm Aut} ( V\times 1)\cong {\rm Aut} \hspace{.5pt} U$ on this subsystem of $U$. \end{proposition} \begin{proof} Let $h\in {\rm Aut}\hspace{.5pt} U$. We must show that $h$ is induced by some element of ${{\rm Aut} ( V\times 1)}\le {\rm Aut} \hspace{.5pt} U$ (cf. ~\ref{p4}). Proposition~\ref{maximal}(i) states that the subsystems $Y_v$ are uniquely determined for $U$. Then Proposition~\ref{maximal}(ii) states that the subsystems $V_{V,f}$ are also uniquely determined for $U$. It follows that $h$ sends $ V\times1=V_{V,1}$ to $V_{ V, f}\subseteq V\times A_6$ for some $f\colon V\to A_6$, and $h$ permutes the subsystems $ Y_v$. Since $h$ sends $X=\bigcap_{v\in V}Y_v$ to itself it also sends $U-X= V\times A$ to itself. In view of ~\ref{M3}, restricting $h$ to the first component in $ V\times A$ induces an isomorphism $ \bar h\colon \!V\to V$; by ~\ref{p4}, \vspace{1pt} $ \bar h$ is also induced by some $g\in {\rm Aut} ( V\times 1)\le {\rm Aut} \hspace{.5pt} U$. Then $ \bar h \bar g^{-1} =1$ on $V$. We will prove that $h=g$. Replace $h$ by $hg^{-1}$, so $ \bar h=1$ on $V$. {\em The remainder of the proof consists of showing that $h=1$.} \vspace{2pt} Since $ (v\times A)^h= v^{\bar h}\times A = v\times A $ and $( V\times 1)^h=V_{ V, f}$ we have $(v,1)^h = (v,f(v))$ for all $v\in V$. Since $h$ permutes the subsystems $ Y_v$, from $(v, 1),(v,1)^h= (v,f(v))\in Y_v $ it follows that $h$ leaves invariant every $ Y_v$. Let $(v,a)^h=(v,f_v(a))$ where $a,f_v(a)\in A$. Then $(v,f_v(1))=(v,1)^h = (v,f(v)) $. Let $b_v\coloneqq f_v(1) =f (v)\in A_6$. {\em We will show that $h$ acts on $V\times A$ by} \begin{equation} \label{va} \mbox{$(v,a)^h=(v, b_va^\alpha)\,$ for all $v\in V, a\in A,$ and some $\alpha\in {\rm Aut} A.$} \end{equation} Let $v_1,v_2,v_3$ be a triple of $ V$. Whenever $a_1a_2a_3=1$, $a_i\in A$, by ~\ref{M3} we obtain a triple $(v_1,a_1), $ $(v_2,a_2), $ $(v_3,a_3)$ and hence also its image under $h$: the triple $(v_1,f_{v_1}(a_1)),$ $ (v_2,f_{v_2}(a_2)),$ $ (v_3,f_{v_3}(a_3))$, so $f_{v_1}(a_1)f_{v_2}(a_2)f_{v_3}(a_3)=1$. Then $$ \mbox{$f_{v_1}(a_1)f_{v_2}(a_2)f_{v_3}(a_1^{-1}a_2^{-1})=1$ \ for all $a_1,a_2\in A$.} $$ Let $a_1=1$ and deduce that $f_{v_2}(a_2) =b_{v_1}^{-1}f_{v_3}(a_2^{-1})^{-1}$; while $a_2=1$ yields $f_{v_1}(a_1) =b_{v_2}^{-1}f_{v_3}(a_1^{-1})^{-1}$. Then $b_{v_1}b_{v_2}b_{v_3}=f_{v_1}( 1)f_{v_2}(1)f_{v_3}(1)=1$ and (after replacing $a_i^{-1}$ by $a_i$) $$ b_{v_3} f_{v_3}(a_1 a_2 )= f_{v_3}(a_1 ) f_{v_3}( a_2 ). $$ Now $b_{v_3}^{-1} f_{v_3}(a_1 a_2 )= b_{v_3}^{-1}f_{v_3}(a_1 ) b_{v_3}^{-1}f_{v_3}( a_2 )$, so that $f_{v_3}(a_1)=b_{v_3}a_1^\alpha$ for some $\alpha\in {\rm Aut} A$ and all $a_1\in A$. Moreover, ${f_{v_2}(a_2) =b_{v_1}^{-1}f_{v_3}(a_2^{-1})^{-1} = b _{v_2}a_2^\alpha}$: we have the same automorphism $\alpha$ for all $v\in V$. This proves \eqref{va}. % \smallskip By \eqref{va} and~\ref{M2}, if $v\in V$ then $a\mapsto b_va^\alpha$ is an automorphism of the partial Steiner triple system $Y-X$. By ~\ref{p6}, $\alpha=1$ and $b_v=1$ for all $v$. Then $h=1$ on $V\times A$. Since every point of $X$ is in a triple containing two points of $Y-X$, it follows that $h=1$, as claimed. \end{proof} % \newpage \subsection{Completion of proof} \label{Completion of proof} In ~\ref{p1}--\ref{p4} we provided the ingredients for the construction of an STS $U$ using Section~\ref{XYV}. Proposition~\ref{Aut A} determined Aut\hspace{.5pt}$U$. Moreover, by~\ref{p1}(a) and Proposition~\ref{Aut A}, $U$ has Aut\hspace{.5pt}$U$-invariant subsystems $V\times 1\supset V_{\scriptscriptstyle\star}\times 1$ such that Aut\hspace{.5pt}$U \cong {\rm Aut} (V\times 1) \cong {\rm Aut}( V_{\scriptscriptstyle\star}\times 1)$ and Aut\hspace{.5pt}$U$ induces $ {\rm Aut}(V\times 1) $ and $ {\rm Aut} ( V_{\scriptscriptstyle\star}\times 1) $ on the respective subsystems. Lemma~\ref{Existence2} states that we have dealt with all admissible $u> 800 ^2v^7$. \qed % \newpage % \rema \rm \begin{Remark} \label{bounds} {\bf Bounding $N_{V_{\scriptscriptstyle\star}} $}. In Lemma~\ref{Existence1} we had $u> 800 ^2v^7$, but the integer $v=|V|$ obtained in Corollary~\ref{=3} is much larger than $|V_{\scriptscriptstyle\star} |$. By Proposition~\ref{VxD}(iv), $|{V_{\scriptscriptstyle\star}}{\bm\times} D |$ is $O(|V_{\scriptscriptstyle\star} |^5)$, so $v$ is $O(|V_{\scriptscriptstyle\star} |^{5\cdot 5})$. Thus, $ N_{V_{\scriptscriptstyle\star}} $ is $O(|V_{\scriptscriptstyle\star} |^{25\cdot 7})$.% \medskip \noindent {\bf Bounding $M_G $}. In \cite{Me} and \cite{Ka} an STS\ $V_{\scriptscriptstyle\star}$ is constructed for which $G\cong{\rm Aut}V_{\scriptscriptstyle\star}$ and $|V_{\scriptscriptstyle\star} |$ is $2^{O(|G|)}$. By the preceding paragraph the same is true for $M_G $. {\em This bound for $M_G$ is ridiculously weak. It seems likely that $M_G$ is polynomial in $|G|,$ but entirely new methods would be needed to prove that. } %\rm \end{Remark} \begin{Remark} \label{comparison} The argument in \cite{Ca} depended on using pairs $X\subset Y$ provided by \cite{DW}, essentially for all possible $x=|X|$ and $y=|Y|$ for which $y\ge 2x+1$. The argument used here only had access to a more limited choice~\ref{p2}(a) of orders $y$ (cf. Lemma~\ref{direct product}(ii)). In \cite{Ca} first $y-x$ was dealt with, at which point $x$ and $y$ were uniquely determined for given $v$ and $u$. This approach can be used in our situation when $u\equiv -1$~(mod~4) but not when $u\equiv 1$~(mod~4). Therefore we have started with a restricted choice of $y$, and then $x$ is uniquely determined for given $v$ and $u$ (Lemmas~\ref{Existence1} and \ref{Existence2}). Our problem was to have a suitably geometric $ Y$ of order $y$ with a subsystem of the required order $x\le (y_0-1)/2< (y-1)/2$. \end{Remark} %\newpage \section {Partial Steiner triple systems} \label {Proof PSTS} \subsection {Theorem~\ref{PSTS}} \label {Proof of Theorem {PSTS}} We first note how our approach differs from that of Cameron \cite{Ca}. He observes: ``In the construction used to prove Theorem 1, if the subsystem contains no triples, its automorphism group is the symmetric group $S_u$, while that of the embedding system is the general linear group $GL(u-1,2)$.'' In other words, the PSTS (partial Steiner triple system) might have too few triples. The first part of our proof eliminates this possibility (cf. Lemma~\ref{V'}(v)). \begin{definition} \label{paths} Let the PSTS $Q_k(x), k\ge 2,$ have the following triples (using two ``paths'' of $k$ triples in the first two rows and an additional point $2k+1$): $$ \begin{array}{lllllllllll} x,1,2 &2,3,4\hspace{28.45pt} \ 4,5,6\hspace{18.9pt}\dots \quad2k-2,2k-1,2k \\ x',1',2' &2',3',4' \qquad \ 4',5',6'\quad\dots \quad(2k-2)',(2k-1)',(2k)' \\ 2k+1,x,x'\ & 2k+1,i,i' \mbox{\hspace{2.4pt} \ for $1\le i\le 2k$} \ \ \\ x,3 ,(2k)' & x,3',2k \end{array} $$ \end{definition} \begin{Remark} \label{paths2} The following properties of $Q_k(x)$ are straightforward: \begin{enumerate} \item[(1)] $Q_k(x)$ has $4k+3$ points, \item[(2)] every point is in at least two triples$ ,$ \item[(3)] the point $2k+1$ is in $2k+1\ge 5$ triples, $x$ is in four triples and every other point is in at most three triples, \item[(4)]every point is in the union of the triples containing $2k+1$, and \item[(5)]${\rm Aut}Q_k(x)=1$. \end{enumerate} (For (5), every automorphism must fix $x$ and $2k+1$, then also $x'$, $1'$, $1$, $2$, \dots.) \end{Remark} Let $V$ be an $n$-point PSTS as in Theorem~\ref{PSTS}. We may assume that $n\ge2$. \begin{Lemma} \label{V'} There is a PSTS\, $V'$ such that \begin{enumerate} \item[(i)] $ V $ is an ${\rm Aut} \hspace{.5pt} V'$-invariant subsystem of $V',$ \item[(ii)]${\rm Aut}\hspace{.5pt} V'$ induces ${\rm Aut}\hspace{.5pt} V$ on $V, $ \item[(iii)]${\rm Aut} \hspace{.5pt} V'\cong{\rm Aut}\hspace{.5pt} V,$ \item[(iv)] $n'\coloneqq |V'|\ge 22,$ and \item[(v)] every point of $V'$ is in at least two triples of $V'.$ \end{enumerate} \end{Lemma} \begin{proof} For every point $x$ of $V$, attach $Q_{n }(x)$ to $V $ so that $V \cap Q_{n}(x) = x$ and the $n$ PSTSs $Q_{n}(x)$ are pairwise disjoint. The union of $V$ and these PSTSs (also using the union of their sets of triples) is a new PSTS $\,V' $ having $n'$ points, where $n' = n|Q_{n}(x)| = n(4n +3) \ge 22 $, which proves (iv). Condition (i) is clear, (v) holds in $V'$ by Remark~\ref{paths2}(2), and (ii) follows from the fact that all $Q_{n }(x)$ are isomorphic and are pairwise disjoint. It remains to prove (iii). By Remark~\ref{paths2}(3), any subsystem $Q$ of $V'$ isomorphic to $Q_n(x)$ has a point $z$ in $2n+1$ triples of $Q$. Since \vspace{2pt} $V'=\bigcup_{x\in V}Q_n(x)$, again by Remark~\ref{paths2}(3) each point of $V'$ is either in $2n+1$ triples of $V'$, at most 3 triples, or (for points of $V$) between 4 and $4+(n-1)/2 <2n+1$ triples. Then $z\notin V$ and $z\in Q_n(x)$ for a unique $x$. By Remark~\ref{paths2}(4), the union of the triples of $V'$ containing $z$ is both $Q$ and $Q_n(x)$, so $Q=Q_n(x)$. Thus, $V'$ determines the points of $V$, any element of ${\rm Aut } V'$ induces an element of ${\rm Aut } V,$ and this yields a homomorphism from ${\rm Aut } V'$ onto ${\rm Aut } V.$ Its kernel fixes every point $x$ of $V$, and hence is 1 by Remark~\ref{paths2}(5). \end{proof} In the rest of the proof we ignore $V$ and work only with $V'$. By Theorem~\ref{main}, \emph{it suffices to construct one STS\, $U$ having $ V '$ as an ${\rm Aut} \hspace{.5pt}U$-invariant subsystem such that ${\rm Aut} \hspace{.5pt}U\cong{\rm Aut}\hspace{.5pt} {V' }$ and ${\rm Aut}\hspace{.5pt} U$ induces ${\rm Aut}\hspace{.5pt} V'$ on $V'$.} As in \cite{Ca}, we use the projective space $P=PG(n'-1,2)$ whose points are the $2^{n'} - 1$ nonempty subsets of (the set of points of) $V' $, the lines of $P$ being all triples of subsets of $ V'$ whose symmetric difference is empty. Any permutation of the points of $V'$ extends uniquely to an automorphism of $P.$ Every point $w$ of $P$ has size $|w|$ as a subset of~$V'$. Again as in \cite{Ca}, we construct from the STS $P$ and the PSTS $V'$ an STS $U$ whose points are those of $P$, as follows: for every triple $a, b, c $ of $V'$, replace the triples \begin{equation} ab, ac, bc \qquad a,b,ab \qquad a,c,ac \qquad b,c,bc \end{equation} of $P$ by the new triples \begin{equation} \label{triples in S} a, b, c \qquad a,ab,ac\qquad b,ab,bc \qquad c,ac,bc \end{equation} (by abuse of notation, we write $a$ and $ab$ for $\{a\}$ and $\{a, b\}$, respectively). This produces a new STS $\,U$, because the new triples cover exactly the same pairs of points as the old ones. Note that \begin{equation} \label{2} \mbox{\em Every point $ab$ is in at most two triples of $U$ that are not lines of $P,$} \end{equation}and that ${\rm Aut } V'$ induces a subgroup of ${\rm Aut } \hspace{.5pt}U$ (as in \cite[p.~468]{Ca}). Moreover, \begin{equation} \label{block is line} \mbox{\em A line of $P$ is also a triple of $U$ if it contains a point $w$ with $|w|>2$.} \end{equation} \begin{Lemma} \label{triples} The lines of $P$ can be determined using the triples in $U$. \end{Lemma} \begin{proof} We will recover the line $\langle x,y\rangle$ of $P$ determined by any given distinct points $x,y$ of $U$. For every point $p $ of $U $ not in the triple of $U$ determined by $x$ and $y$ there are distinct triples $$p,x,x_1 \qquad p,y,y_1 \qquad x_1,y_1,z \qquad p,z,q$$ of $U$, producing a 7-set $U(p,x,y) \coloneqq \{p,x,y,x_1,y_1,z,q\}$ of points of $U$. There are at least $(2^{n'-2}-1) -n'-{\binom{n'}{2}}$ planes \vspace{3pt} of $P$ containing $x$ and $y$ but containing no point $w$ of $P$ with $|w|\le 2$. Every point $w\notin \langle x,y\rangle$ in such a plane has $|w|>2$; by \eqref{block is line}, every such plane has the form $U(p,x,y)$ for any of its four points $p\notin \langle x,y\rangle$. Thus, by Lemma~\ref{V'}(iv), if $p$ is one of at least $4(2^{n'-2}-1-n'-{\binom{n'}{2}})>\frac{3}{4}|U|$ points in the union $S $ of these planes but not in $\langle x,y\rangle$, then \begin{itemize} \item [\rm(i)] every set $U(p,x,y)$ occurs for at most four points $p$, and \item[\rm(ii)] distinct sets $U(p,x,y)$ have the same intersection of size 3. \end{itemize} (The intersection in (ii) is the line $\langle x,y\rangle=\{x,y,z\}$.) If $S '$ is another set of more that $\frac{3}{4}|U|$ points satisfying (i)-(ii), then $|S\cap S' | \ge\frac{3}{4}|U|+\frac{3}{4}|U|- |U|=\frac{1}{2}|U|=\frac{1}{2}(2^{n'}-1) \ge \frac{1}{2}(2^{22}-1) $, and hence by (i) $ S\cap S'$ contains distinct sets $U(p,x,y)$. Those sets produce the same set of size 3 in (ii). Thus, we have obtained the line $ \langle x,y\rangle$ of $P$ using the triples of $U$. \end{proof} %\smallskip\smallskip Note that we have not yet used Lemma~\ref{V'}(v). \begin{proposition} ${\rm Aut}\hspace{.5pt} U\cong {\rm Aut} V'\cong {\rm Aut} V.$ \end{proposition} \begin{proof} We now have the triples in $U$ and the triples in $P$. Let $T$ denote the set of triples of $U$ that are not triples in $P$ (these are the triples in~\eqref{triples in S}, and hence consist of points such as $a\in V'$ or $ab$). Let $x \in U$. \begin{itemize} \item [\rm(1)] If $x$ is in at least four triples in $T,$ then $x\in V'$ by \eqref{2}. \item[\rm(2)] Every point of $V'$ is in at least four triples in $T,$ by Lemma~\ref{V'}(v) and~\eqref{triples in S}. \end{itemize} Thus {\em$U$ uniquely determines $V',$} so ${\rm Aut } \hspace{.5pt} U$ induces a subgroup of ${\rm Aut }\hspace{.5pt} V'.$ Since ${\rm Aut } V'$ induces a subgroup of ${\rm Aut } \hspace{.5pt}U,$ by Lemma~\ref{V'}(iii) we have ${\rm Aut }\hspace{.5pt} U \cong {\rm Aut } \hspace{.5pt} V' \cong {\rm Aut } V,$ and we are done. \end{proof} %{\noindent\em Proof of {\rm Theorem~\ref{PSTS}}.} \begin{proof}[{Proof of Theorem~\ref{PSTS}}] We have embedded the original PSTS $V$ into an STS $\,U$ such that ${\rm Aut }\hspace{.5pt} U $ leaves $V$ invariant, induces ${\rm Aut } V $ on $V$, and is isomorphic to ${\rm Aut } V $. Now apply Theorem~\ref{main} to $U$ (in place of $V$) in order to obtain STSs behaving as in Theorem~\ref{PSTS}. \end{proof} \subsection{Corollaries} \label{corollaries} We note some consequences of Theorem~\ref{PSTS}. We will use a natural graph on the points of a PSTS $W$, with two points joined if they are in a triple. If this graph is not connected we can embed $W$ in an arbitrarily large STS, which is clearly connected (preservation of the automorphism group is even possible by Theorem~\ref{PSTS}, but this will not be needed). \begin{corollary} \label{2 STS} Given partial Steiner triple systems $V$ and $W,$ there is an integer $N_{V,W}$ such that$,$ for each admissible $u\ge N_{V,W} ,$ there is a Steiner triple system $U$ on $u$ points having a subsystem $W'\cong W$ and an ${\rm Aut}\hspace{.5pt} U$-invariant subsystem $V'\cong V$ with $ W' \cap V'=\emptyset$ such that ${\rm Aut}\hspace{.5pt} U \cong{\rm Aut} V'$ and ${\rm Aut}\hspace{.5pt} U $ induces ${\rm Aut} V'$ on~$V'$. \end{corollary} \begin{proof} Here $V$ and the desired $U$ are as usual, the new aspect is to include $W$ as well; we have no control over the PSTS $V\dot \cup W$. By the preceding remarks, we may assume that $W$ is an STS and hence is connected, and that $W\cap V=\emptyset$, $n=|W|\ge2$ and $n>|V|$. Let $x_1,\dots,x_n$ be the points of $W$. Then Definition~\ref{paths} applies with $k=i+n\ge 3$; attach pairwise disjoint PSTSs $Q_{i+n}(x_i) $ to $W$ in such a way that $Q_{i+n}(x_i) \cap W =x_i$ for every $i$. The union of $W$ and all $Q_{i+n} (x_i) $ is a connected PSTS $\widehat W$. Every $Q_{i+n}(x_i)$ has a unique point in $2(i+n)+1\ge 2n+3 >\frac{1}{2}(n-1)+4$ triples (by Remark~\ref{paths2}(3)), and $\widehat W$ has no other such points ($x_i$ is in at most $\frac{1}{2}(n-1)+4$ triples of $\widehat W$, again by Remark~\ref{paths2}(3)). Then $W$ can be recovered from $\widehat W$ using Remark~\ref{paths2}(3)-(4). The PSTSs \vspace{1pt} $Q_{i+n}(x_i), 1\le i\le n, $ have different orders, so from Remark~\ref{paths2}(5) it follows that ${\rm Aut} \widehat W=1$. Since $|\widehat W|>|V|$ and $\widehat W$ is a connected component of the graph on the disjoint union $\widehat W\dot{\cup} V'$ of $\widehat W $ and $ V'\cong V$,\, ${\rm Aut} (\widehat W\dot{\cup} V')$ leaves $\widehat W$ invariant and hence acts on $V'$. Then ${\rm Aut} (\widehat W\dot{\cup} V')\cong {\rm Aut} V'$, so apply Theorem~\ref{PSTS} to $\,\widehat W\dot{\cup} V'$. \end{proof} %\medskip The first step in the above proof was to embed an arbitrary STS into one whose automorphism group is trivial. This suggests a strengthening of Theorem~\ref{corollary}: \begin{corollary} \label{corollary2} If $V _1,\dots,V_m $ are partial Steiner triple systems and $G$ is a finite group$,$ then there is an integer $N_{V _1,\dots,V_m,G}$ such that$,$ for each admissible $u\ge N_{V _1,\dots,V_m,G},$ there is a Steiner triple system $U$ on $u$ points such that $V _1,\dots,V_m $ are isomorphic to pairwise disjoint subsystems of $U$ and ${\rm Aut} \hspace{.5pt}U\cong G.$ \end{corollary} \begin{proof} Let $V$ be an STS with ${\rm Aut} \hspace{.5pt}V\cong G $ \cite{Me}. Apply the preceding corollary to $V$ and $W$, where $W$ is the disjoint union of (copies of) $V _1,\dots,V_m $. \end{proof} \begin{corollary} \label{2 groups} If $G$ and $H$ are finite groups then there is an integer $N _{G,H}$ such that$,$ for each admissible $u\ge N _{G,H},$ there is a Steiner triple system $U$ on $u$ points having a subsystem $W$ such that ${\rm Aut}\hspace{.5pt} U\cong G $ and ${\rm Aut} \hspace{.5pt} W\cong H $. \end{corollary} \begin{proof} Let $V$ and $W$ be STSs such that ${\rm Aut} \hspace{.5pt} V\cong G $ and ${\rm Aut} W\cong H $ \cite{Me}. Apply Corollary~\ref{2 STS} to the pair $V,W$ in order to obtain an STS $\,U$ behaving as stated. \end{proof} %\medskip This corollary can be iterated in two ways: one involves disjoint subsystems with arbitrary given automorphism groups; another involves a nested sequence of subsystems with arbitrary given automorphism groups. Our final corollary concerns retaining a {\em subgroup} of the automorphism group of an STS but not the full automorphism group. Notation: If $G$ is a group acting on a set $X$, and if $Y\subset X$, then the \emph{set-stabilizer} $G_Y$ is $\{g\in G \mid \mbox{$g$ sends $Y $ to itself\}}$. (In the corollary $V_1$ need not be ${\rm Aut} \hspace{.5pt} V $-invariant, and ${\rm Aut}\hspace{.5pt} V_1 $ need not be a subgroup of ${\rm Aut} \hspace{.5pt} V $.) \begin{corollary} \label{subSTS} If $V_1 $ is a subsystem of order $>1$ of a partial Steiner triple system $V ,$ then there is an integer $N'_{V,V_1}\!$ such that$,$ for each admissible $u\ge N'_{V,V_1},$ there is a Steiner triple system $U$ on $u$ points having $V $ and $V_1$ as ${\rm Aut} \hspace{.5pt}U$-invariant subsystems such that ${\rm Aut}\hspace{.5pt} U \cong ({\rm Aut}\hspace{.5pt} V)_{V_1}$ and ${\rm Aut}\hspace{.5pt} U$ acts on $V$ as $({\rm Aut}\hspace{.5pt} V)_{V_1}$. \end{corollary} \begin{proof} First we replace $V$ by an STS: use Theorem~\ref{PSTS} to find an STS $\,\hat V $ containing $V$ such that $ {\rm Aut}\hat V$ leaves $V$ invariant, induces ${\rm Aut} V$ on $V$ and is isomorphic to ${\rm Aut} V$. Then $ ({\rm Aut}\hat V)_{V_1} \cong ({\rm Aut}\hspace{.5pt} V)_{V_1}$. Let $z$ be a new point, and let $x\mapsto x'$ be a bijection from $V_1$ to a set $V_1'$ disjoint from $\hat V\cup \{ z\}; $ this bijection turns $V_1'$ into a PSTS. Form a PSTS\, $W, $ with $\hat V\cup \{ z\}\cup V_1' $ as its set of points, by using the triples in $\hat V\cup V_1' $ and \vspace{1pt} including a new triple $x,z,x'$ for every $x\in V_1$. Every $g$ in $ ({\rm Aut}\hat V)_{V_1}$ acts as an automorphism of $W$ via $z^g=z$ and $(x')^g = (x^g)'$ for $x\in V_1$. The set $V_1$ is uniquely determined as the set of points of $W$ lying in triples~with the maximal number of other points (namely, $(|\hat V|-1)+2$ points); $\hat V -V_1$ is uniquely determined as the set of points of $W$ lying in triples with exactly $|\hat V| -1$ points. Then ${\rm Aut}\hspace{.5pt} W$ induces $ ({\rm Aut} \hspace{.5pt} \hat V)_{V_1}$ on $\hat V$. Let $K$ denote the pointwise stabilizer of $\hat V$ in ${\rm Aut} \hspace{.5pt} W$. For distinct $x,y\in V_1$, the triples $x,z,x'$ and $y,z,y'$ meet at $z$, so $K$ fixes $z$ and then also all points of $V_1'$. Thus, $K=1$ and ${\rm Aut}\hspace{.5pt} W=({\rm Aut}\hspace{.5pt} \hat V)_{V_1}$. Now apply Theorem~\ref{PSTS} to $W$. \end{proof} \begin{Remark}\em If $G$ and $H$ are finite groups with $G>H,$ then there is an integer $N '_{G,H}$ such that$,$ for each admissible $u\ge N _{G,H},$ there a Steiner triple system $U$ on~$u$ points having a subsystem $W$ such that ${\rm Aut} \hspace{.5pt}U\cong G $ and $({\rm Aut}\hspace{.5pt} U)_W \cong {\rm Aut} W\cong H.$ \rm The proof involves a few straightforward changes in \cite[Sec.~2]{Ka}, which we briefly outline. {\bf 1.} Let $\Gamma$ be an $n$-vertex connected graph having a connected induced subgraph $\Gamma'$ such that ${\rm Aut} \Gamma \cong G$, \,$G$ acts semiregularly on the vertices of $\Gamma$, and $({\rm Aut} \Gamma)_{\Gamma'} \cong {\rm Aut} \Gamma' \cong H$. (Use the standard colored Cayley graphs for $G$ and $H$ and replace colored edges by suitable graphs.) {\bf 2.} Consider the vector space $V$ in Section~\ref{Increasing $V$.}. In order to conform to the notation in \cite[Sec.~2]{Ka} let $V_F$ and $V$ denote this as an $F$-space and as an ${\bf F}_2$-space, respectively. Assume that $G$ acts on the basis of $V_F$ as it does on the vertices of $\Gamma$. Let $V'_F$ denote the $F$-span of the vertices of $\Gamma'$, and let $V'$ be $V'_F$ viewed as an ${\bf F}_2$-space. {\bf 3.} In \cite[Sec.~2]{Ka} there is a construction of an STS $\,U$ with ${\rm Aut} \hspace{.5pt} U \cong G$, using $V_F$, $V$ and $\Gamma$, and two auxiliary STSs on 15 points. Restricting the construction to $V_F'$, $V'$ and $\Gamma'$ produces a subsystem $U'$ of $U$ obtained using these ingredients in the same manner that $U$ was. In particular, $({\rm Aut} \hspace{.5pt} U)_{U'} \cong {\rm Aut} \hspace{.5pt}U' \cong H$, as required. \end{Remark} \longthanks{ We are grateful to a perspicacious, precise and patient referee, whose comments led to significant improvements.} \bibliographystyle{mersenne-plain} \bibliography{ALCO_KANTOR_216} \end{document}