Skip to main content

Math 344-1,2: Kursobjekt

Aaron Greicius

Section 2.13 Free products

Definition 2.13.1. Free product of groups.

A free product of a family of groups \(\{G_\alpha\}_{\alpha\in I}\) is a pair \((G, \{i_\alpha\colon G_\alpha\rightarrow G\})\) consisting of a group \(G\) and a family of group homomorphisms \(i_\alpha\colon G_\alpha\rightarrow G\) satisfying the following universal mapping property:
  • Universal mapping property.
    If \(H\) is any group, and if \(\{\phi_\alpha\colon G_\alpha\rightarrow H\}_{\alpha\in I}\) is any family of group homomorphisms, then there is a unique group homomorphism \(\phi\colon G\rightarrow H\) satisfying
    \begin{equation*} i_\alpha\circ\phi=\phi_\alpha \end{equation*}
    for all \(\alpha\in I\text{.}\)
    Commutative diagram for free products
As we will see, such a group \(G\) is unique up to isomorphism. We will write \(\prod_{\alpha\in I}^*G_\alpha\) to denote a free product of \(\{G_\alpha\}_{\alpha\in I}\text{.}\) When \(I=\{\alpha_1,\alpha_2,\dots, \alpha_n\}\) is finite, we write \(G_{\alpha_1}*G_{\alpha_2}*\cdots *G_{\alpha_n}\) to denote a free product of the \(G_{\alpha_k}\text{.}\)

Proof.

  1. Fix \(\alpha\in I\text{.}\) We use the universal mapping property, setting \(H=A_{\alpha}\text{,}\) letting \(\phi_{\alpha'}\colon A_{\alpha'}\rightarrow A_\alpha\) be the zero map for all \(\alpha'\ne \alpha\text{,}\) and letting \(\phi_\alpha\colon A_\alpha\rightarrow A_{\alpha}\) be the identity map \(\phi_\alpha=\id_{A_{\alpha}}\text{.}\) We conclude there is a map \(\phi\colon G\rightarrow A_\alpha\) satisfying \(\id_{A_\alpha}=\phi\circ i_\alpha\text{.}\) Since \(\id_{A_\alpha}\) is injective, the map \(i_\alpha\) must be injective.
  2. The proof here is very similar to the one from Theorem 2.13.2. Since \(G\) is a free product, the maps \(i_\alpha'\colon G_\alpha\rightarrow G'\) give rise to a homomorphism \(\phi\circ G\rightarrow G'\) satisfying
    \begin{equation*} i_\alpha'=\phi\circ i_\alpha \end{equation*}
    for all \(\alpha\in I\text{.}\) And since \(G'\) is a free product, the maps \(i_\alpha\colon G_\alpha\rightarrow G\) give rise to a homormorphism \(\psi\colon G'\rightarrow G\) satisfying
    \begin{equation*} i_\alpha=\psi\circ i_\alpha' \end{equation*}
    for all \(\alpha\in I\text{.}\) It is then straightforward to show that
    \begin{align*} (\psi\circ\phi)\circ i_\alpha \amp = i_\alpha\\ (\phi\circ\psi)\circ i_{\alpha'} \amp =i_{\alpha'} \end{align*}
    for all \(\alpha\in I\text{,}\) from whence it follows from the uniqueness condition of the universal mapping property that
    \begin{align*} \psi\circ\phi \amp = \id_G\\ \phi\circ\psi \amp =\id_{G'}\text{.} \end{align*}
We now endeavor to show that free products of groups exist. Our construction is fairly concrete in the end: essentially, we will build a group \(G\cong \prod\limits_{\alpha\in I}^* G_\alpha\) whose elements are certain types of words built from letters ranging over the elements of the groups \(G_\alpha\text{.}\) We make this precise below.

Definition 2.13.3. Words and reduced words.

Let \(\{G_\alpha\}_{\alpha\in I}\) be a family of groups, and let \(\mathcal{A}=\coprod G_\alpha\) be the disjoint union of their underlying sets. Given an integer \(n\geq 0\text{,}\) a word of length \(n\) on the alphabet \(\mathcal{A}\) is an \(n\)-tuple \(w=(g_{\alpha_1},g_{\alpha_2},\dots, g_{\alpha_n})\) where \(g_{\alpha_k}\in G_{\alpha_k}\) for all \(1\leq k\leq n\text{.}\) The empty word is the empty sequence \(()\text{,}\) the unique tuple on \(\mathcal{A}\) of length 0.
A word \(w\) is reduced if either \(w=()\) or \(w=(g_{\alpha_1},g_{\alpha_2},\dots, g_{\alpha_n})\text{,}\) where \(g_{\alpha_k}\ne e_{\alpha_k}\) for all \(1\leq k\leq n\) and \(\alpha_{k}\ne \alpha_{k+1}\) for all \(1\leq k\leq n-1\text{.}\)

Proof.

Let \(G\) be the set of all reduced words on the alphabet \(\mathcal{A}=\coprod G_\alpha\text{.}\) We wish to define a group operation on \(G\text{.}\) A natural guess for an operation would be sequence concatenation; however, the concatenation of two reduced words is not necessarily a reduced word. This is easily corrected by defining our group operation \(*\colon G\times G\rightarrow G\) recursively as follows.
  • For any \(w\in G\text{,}\) define \(()*w=w*()=w\text{.}\)
  • Given \(w=(g_{\alpha_1},g_{\alpha_2},\dots, g_{\alpha_n})\) and \(u=(h_{\beta_1}, h_{\beta_2},\dots, h_{\beta_m})\) with \(n,m\geq 1\text{,}\) define
    \begin{equation*} w*u=\begin{cases} (g_{\alpha_1},\dots, g_{\alpha_n},h_{\beta_1},\dots, h_{\beta_m}) \amp \text{if } \alpha_n\ne \beta_1 \\ (g_{\alpha_1},\dots, g_{\alpha_n}h_{\beta_1},\dots, h_{\beta_m}) \amp \text{if } \alpha_n=\beta_1 \text{ and } h_{\beta_1}\ne g_{\alpha_n}^{-1} \\ (g_{\alpha_1},g_{\alpha_2},\dots, g_{\alpha_{n-1}})*(h_{\beta_1}, h_{\beta_2},\dots, h_{\beta_{m-1}}) \amp \text{if } \alpha_n=\beta_1 \text{ and } h_{\beta_1}=g_{\alpha_n}^{-1}. \end{cases} \end{equation*}
It is easy to show by induction on the maximum length of \(w\) and \(u\) that \(w*u\) is reduced. However, it is quite difficult to show that \(*\) is actually a group operation directly! Specifically, the associative property is somewhat of a nightmare to verify. We will do so indirectly, using what is often called the trick of van der Waerden.

Proof that \(*\) is a group operation.

It is easy to see that the trivial word \(()\) is an identity element with respect to \(*\text{.}\) Furthermore, if \(w=(g_1,g_2,\dots, g_n)\) is a reduced word of length \(n\geq 1\text{,}\) the word \(w^{-1}=(g_n^{-1}, \dots, g_1^{-1})\) clearly satisfies \(w^{-1}u=()\) if and only if \(u=w\text{.}\) It follows that every element of \(G\) has a two-sided inverse with respect to \(*\text{.}\) The only diffficult thing to show is that \(*\) is associative, as mentioned above. This is where the trick of van der Waerden comes in.
Let \(S_G\) be the set of all permutations of \(G\text{.}\) In other words, \(S_G\) is the set of all bijections \(\sigma\colon G\rightarrow G\text{:}\) a group under function composition. We will define an injection \(i\colon G\hookrightarrow S_G\) that satisfies \(i(w*u)=i(w)\circ i(u)\) for all reduced words \(w,u\in G\text{.}\) Since composition \(\circ\) is an associative operation on \(S_G\text{,}\) it will then follow that \(*\) is an associative operation on \(G\text{.}\)
Before getting to the function \(i\text{,}\) we first define for any \(g\in G_\alpha\) a map \(\sigma_g\colon G\rightarrow G\) as follows.
  1. \(\displaystyle \sigma_e=\id_G\)
  2. \(\sigma_g(())=(g)\) if \(g\ne e\text{.}\)
  3. \(\sigma_g((g_{\alpha_1},g_{\alpha_2},\dots, g_{\alpha_n}))= (g,g_{\alpha_1},g_{\alpha_2},\dots, g_{\alpha_n})\) if \(g\ne e\) and \(\alpha\ne \alpha_1\text{.}\)
  4. \(\sigma_g((g_{\alpha_1},g_{\alpha_2},\dots, g_{\alpha_n})) = (gg_{\alpha_1},g_{\alpha_2},\dots, g_{\alpha_n})\) if \(g\ne e\) and \(\alpha=\alpha_1\) and \(g\ne g_{\alpha_1}^{-1}\text{.}\)
  5. \(\sigma_g((g_{\alpha_1},g_{\alpha_2},\dots, g_{\alpha_n}))= (g_{\alpha_2},\dots, g_{\alpha_n})\) if \(g\ne e\) and \(\alpha=\alpha_1\) and \(g=g_{\alpha_1}^{-1}\text{.}\)
It is now a straightforward exercise to show that if \(g,h\in G_{\alpha}\text{,}\) then \(\sigma_g\circ \sigma_h=\sigma_{gh}\text{.}\) Indeed, this is obvious if \(g=e\) or \(h=e\text{;}\) otherwise, one shows directly that \(\sigma_g(\sigma_h(w))=\sigma_{gh}(w)\) for any reduced word \(w=(g_1,g_2,\dots, g_n)\text{,}\) treating cases (2)-(5) in the definition above separately. For example, if \(g_1=h^{-1}\in G_\alpha\text{,}\) then we have
\begin{align*} \sigma_g(\sigma_h(w)) \amp =\sigma_g(g_2,\dots, g_n)=(g,g_2,\dots, g_n) \amp (g_2\notin G_{\alpha})\\ \sigma_{gh}(w) \amp =(ghg_1,g_2,\dots,g_n)=(g,g_2,\dots, g_n) \text{.} \end{align*}
From the property \(\sigma_{g}\circ\sigma_{h}=\sigma_{gh}\) it now follows that \(\sigma_g\) is invertible for any \(g\in G_\alpha\text{.}\) Indeed, we have
\begin{equation*} \sigma_{g}\circ\sigma_{g^{-1}}=\sigma_{g^{-1}}\circ \sigma_g=\sigma_e=\id_G\text{.} \end{equation*}
Thus \(\sigma_g\in S_G\) for all \(g\in G_\alpha\text{.}\)
We now define \(i\colon G\rightarrow S_G\) as follows:
\begin{align*} i(()) \amp =\sigma_e=\id_G\\ i((g_{\alpha_1},g_{\alpha_2},\dots, g_{\alpha_n}) \amp=\sigma_{g_{\alpha_1}}\circ \sigma_{g_{\alpha_2}}\cdots \circ \sigma_{\alpha_{n}}\text{.} \end{align*}
The fact that \(i(w*u)=i(w)*i(u)\) for any reduced words \(w,u\in G\) now follows from our definition of \(*\) and the fact that \(\sigma_g\circ \sigma_h=\sigma_{gh}\) for any \(g,h\in G_{\alpha}\text{.}\)
Lastly, we must show that \(i\) is injective. We first observe that \(i(w)=i(())\) if and only if \(w=()\text{.}\) Indeed, for any nontrivial reduced word \(w=(g_1,g_2,\dots, g_n)\) we have
\begin{equation*} \sigma_{g_{\alpha_1}}\circ \sigma_{g_{\alpha_2}}\cdots \circ \sigma_{g_{\alpha_n}}(())=(g_1,g_2,\dots, g_n)\text{,} \end{equation*}
and thus \(i(w)\ne i(())=\id_G\text{.}\) Now suppose that \(w=(g_1,g_2,\dots, g_n)\) and \(u=(h_1,h_2,\dots, h_m)\) are two nontrivial words with \(i(w)=i(u)\text{.}\) Letting \(u^{-1}=(h_m^{-1}, h_{m-1}^{-1}, \dots, h_1^{-1})\) (see above), we have
\begin{align*} i(w)=i(u) \amp \implies i(u^{-1})\circ i(w)=i(u^{-1})\circ i(u)\\ \amp \implies i(u^{-1}*w)=i(u^{-1}*u)\\ \amp \implies i(u^{-1}*w)=i(())\\ \amp \implies i(u^{-1}*w)=\id_G\text{.} \end{align*}
Since \(i(u^{-1}*w)=i(())=\id_G\text{,}\) it follows from our observation above that \(u^{-1}*w=()\text{,}\) and hence that \(w=u\text{.}\)
Knowing that \(G\) is a group with operation \(*\) as above, we now show that \(G\) is a free product of \(\{G_\alpha\}_{\alpha\in I}\text{.}\)
First, for all \(\alpha\in I\) we define \(i_\alpha\colon G_{\alpha}\rightarrow G\) as follows: \(i_\alpha(e)=()\text{,}\) and \(i_\alpha{g)=(g)\text{,}\) a reduced word of length one, if \(g\ne e\text{.}\) It is easy to see that \(i_\alpha\) is a group homomorphism.
Next we verify that \((G, \{i_\alpha\colon G_{\alpha}\rightarrow G\})\) satisfies the universal mapping property. Given a group \(H\) and family of homomorphisms \(\phi_\alpha\colon G_{\alpha}\rightarrow H\text{,}\) define \(\phi\colon G\rightarrow H\) as follows:
\begin{align*} \phi(()) \amp = e\\ \phi((g_{\alpha_1},g_{\alpha_2},\dots, g_{\alpha_n})) \amp = i_{\alpha_1}(g_{\alpha_1))i_{\alpha_2}(g_{\alpha_2)\cdots i_{\alpha_n}(g_{\alpha_n})\text{.} \end{align*}
It is clear that any homomorphism \(\phi\) satisfying \(\phi_{\alpha}=\phi\circ i_\alpha\) must be so defined, since \(i_{\alpha}(g)=(g)\text{,}\) and since for any reduced word \(w=(g_{\alpha_1},g_{\alpha_2},\dots, g_{\alpha_n})\) we have
\begin{equation*} w=(g_{\alpha_1})*(g_{\alpha_2})*\cdots* (g_{\alpha_n})=i_{\alpha_1}(g_{\alpha_1})*i_{\alpha_2}(g_{\alpha_2})*\cdots *i_{\alpha_n}(g_{\alpha_n)\text{.} \end{equation*}
This proves uniqueness of \(\phi\text{.}\)
It remains only to show that \(\phi\) is in fact a group homomorphism. This is a straightforward, if somewhat tedious exercise of going through the different cases in the recursive description of \(w*u\) above and showing that in each case we have \(\phi(w*u)=\phi(w)*\phi(u)\text{.}\)

Proof.

First we give names to some of the maps our various free products come equipped with.
\begin{align*} i\amp\colon G \rightarrow G*H\\ i'\amp\colon H \rightarrow G*H\\ i_\alpha\amp\colon G_\alpha \rightarrow G\\ i_{\alpha'}\amp\colon G_{\alpha'} \rightarrow H\text{.} \end{align*}
Furthermore for each \(\beta\in I\sqcup J\) we define
\begin{equation*} j_{\beta}=\begin{cases} i\circ i_{\alpha}\amp \text{if } \beta=\alpha\in I\\ i'\circ i_{\alpha'}\amp \text{if } \beta=\alpha'\in J. \end{cases} \end{equation*}
We now show that \((G*H, \{j_\beta\colon G_\beta\rightarrow G*H\})\) is a free product of \(\{G_\beta\}_{\beta\in I\sqcup J}\) by verifying that it satisfies the universal mapping property. Let \(T\) be any group equipped with a collection of homomorphisms \(\{\phi_\beta\colon G_\beta\rightarrow T\}_{\beta\in I\sqcup J}\text{.}\) The subcollections \(\{\phi_\alpha\}_{\alpha\in I}\) and \(\{\phi_{\alpha'}\}_{\alpha'\in J}\) give rise to maps \(\phi\colon G\rightarrow T\) and \(\phi'\colon H\rightarrow T\) making the diagram below commutative.
A commutative diagram
The universal mapping property for \(G*H\) now guarantees a unique map \(\widehat{\phi}\colon G*H\rightarrow T\) that makes the new triangles in the commutative diagram below commutative.
Another commutative diagram
But then we have
\begin{align*} \overline{\phi}\circ j_\alpha \amp =\overline{\phi}\circ i\circ i_\alpha\\ \amp = \phi\circ i_\alpha \amp (\overline{\phi}\circ i=\phi\\ \amp = \phi_\alpha \end{align*}
for all \(\alpha\in I\text{.}\) An identical argument shows, \(\overline{\phi}\circ j_{\alpha'}=\phi_{\alpha'}\) for all \(\alpha'\in J\text{.}\) Lastly, the uniqueness of \(\overline{\phi}\) is guaranteed by its arising uniquely from the maps \(\phi, \phi'\text{,}\) which themselves arise uniquely from the maps \(\phi_\alpha\) and \(\phi_{\alpha'}\text{.}\)

Definition 2.13.6. Least normal subgroup.

Let \(S\) be a subset of the group \(G\text{.}\) The least normal subgroup containing \(S\) is the intersection of all normal subgroups of \(G\) that contain \(S\text{.}\) Equivalently, the least normal subgroup containing \(S\) is the unique normal subgroup \(N\normalin G\) such that (i) \(S\subseteq N\) and (ii) if \(S\subseteq N'\normalin G\text{,}\) then \(N\subseteq N'\text{.}\)
PrevTopNext