free product with amalgamated subgroupFreeProductWithAmalgamatedSubgroup2003-01-30 02:21:172005-01-28 11:22:53Definitionpushout of groups%\documentclass{amsart}
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\begin{defn}
Let $G_k$, $k=0,1,2$ be groups and $i_k\co G_0\to G_i$, $k=1,2$ be monomorphisms.
The free product of $G_1$ and $G_2$ with amalgamated subgroup $G_0$,
is defined to be a group $G$ that has the following
two properties
\begin{enumerate}
\item there are homomorphisms $j_k\co G_k\to G$, $k=1,2$ that make the following
diagram commute
$$\xymatrix{
&{G_1}\ar[dr]^{j_1}&\\
{G_0}\ar[ur]^{i_1}\ar[dr]_{i_2} & & {G}\\
&{G_2}\ar[ur]_{j_2}&
}
$$
\item $G$ is universal with respect to the previous property, that is for
any other group $G'$ and homomorphisms $j_k'\co G_k\to G'$, $k=1,2$ that
fit in such a commutative diagram there is a unique homomorphism $G\to G'$
so that the following diagram commutes
$$\xymatrix{
&{G_1}\ar[dr]^{j_1}\ar@/^1pc/[drr]^{j_1'}& &\\
{G_0}\ar[ur]^{i_1}\ar[dr]_{i_2} & & {G}\ar[r]^{!}&{G'}\\
&{G_2}\ar[ur]_{j_2}\ar@/_1pc/[urr]_{j_2'}& &
}
$$
\end{enumerate}
\end{defn}
It follows by ``general nonsense'' that the free product
of $G_1$ and $G_2$ with amalgamated subgroup $G_0$, if it exists,
is ``unique up to unique isomorphism.'' The free product
of $G_1$ and $G_2$ with amalgamated subgroup $G_0$, is denoted
by $G_1\bigstar_{G_0} G_2$. The following theorem asserts its existence.
\begin{thm}
$G_1\bigstar_{G_0} G_2$ exists for any groups $G_k$, $k=0,1,2$ and
monomorphisms $i_k\co G_0\to G_i$,
$k=1,2$.
\end{thm}
\begin{proof}[\textbf{Sketch of proof}]
Without loss of generality assume that $G_0$ is a subgroup of $G_k$ and
that $i_k$ is the inclusion for $k=1,2$. Let
$$G_k=\langle (x_{k;s})_{s\in S}\, |\, (r_{k;t})_{t\in T} \rangle$$
be a presentation of $G_k$ for $k=1,2$. Each $g\in G_0$ can be expressed as
a word in the generators of $G_k$; denote that word by $w_k(g)$ and let
$N$ be the normal closure of $\{w_1(g)w_2(g)^{-1}\,|\,g\in G_0\}$ in the \PMlinkname{free product}{FreeProduct} $G_1\bigstar G_2$. Define
$$G_1\bigstar_{G_0} G_2:=G_1\bigstar G_2/N\,$$
and for $k=0,1$ define $j_k$ to be the inclusion into the free product
followed by the canonical projection. Clearly (1) is satisfied, while (2)
follows from the universal properties of the free product and the quotient group.
\end{proof}
Notice that in the above proof it would be sufficient to divide by the relations
$w_1(g)w_2(g)^{-1}$ for $g$ in a generating set of $G_0$. This
is useful in practice when one is interested in obtaining a presentation of
$G_1\bigstar_{G_0} G_2$.
In case that the $i_k$'s are not injective the above still goes through
verbatim. The group thusly obtained is called a ``pushout''.
Examples of free products with amalgamated subgroups are provided by Van
Kampen's theorem.