Are free products of finite groups virtually free?

Is the free product $A*B$ of two nontrivial finite groups always virtually free? If yes, is it easy to show?

Solution 1:

Yes, this is a consequence of a theorem due to Nielsen. In fact, one can prove that for any two groups $A$ and $B$ (finite or not) the kernel $F$ of the canonical map $A \ast B \to A \times B$ (which is always a surjection) is freely generated by the set of commutators $aba^{-1}b^{-1}$ with $a \in A \smallsetminus \{1\}$ and $b \in B \smallsetminus \{1\}$.

Given this, we see that if $A$ and $B$ are finite, then $F$ is a free group on $r = (|A|-1)(|B|-1)$ generators of index $[G:F] = |A| |B|$ in $G = A \ast B$.

Note: This confirms the formula using Euler characteristics given in Geoff's answer: $$\chi(A \ast B) = \frac{1}{|A|} + \frac{1}{|B|} - 1 = \frac{|A| + |B| - |A||B|}{|A||B|} = \frac{1-r}{[G:F]}$$

For a proof of Nielsen's theorem see e.g. Lyndon, Two notes on Rankin's book on the modular group, Journal of the Australian Mathematical Society (1973), 16, pp 454–457, Theorem 2. The proof is not terribly difficult, but it is not trivial.

For the convenience of the reader, here's Nielsen's paper A basis for subgroups of free groups, Math. Scand. 3 (1955), 33–45.

Alternatively, see Serre, Arbres, amalgames, $\operatorname{SL}_2$, Proposition 4 in N^o 1.1.3. (or page 6 of the English translation).

Solution 2:

I think it can also be seen using the theory of Euler-Poincare characteristics as developed by C.T.C Wall, and explained in Serre's book "Trees." Excluding the case that either $A$ or $B$ is trivial, the key point to note is that $\chi(A*B) = \frac{1}{|A|} + \frac{1}{|B|}-1 \leq 0,$ while if $G = A*B,$ and $G$ has a free group $F$ on $r$ generators, then the general theory gives $\chi(G) = \frac{1-r}{[G:F]}.$ Once there is a free (not necessarily normal) subgroup of finite index, we are done, of course.

Solution 3:

A sketch of proof only based on algebraic topology:

Let $A$ and $B$ be two finite groups, $C_A$ and $C_B$ two complexes such that $\pi_1(C_A) \simeq A$ and $\pi_1(C_B) \simeq B$, $(\widehat{C_A},p_A)$ and $(\widehat{C_B},p_B)$ the universal covering of $C_A$ and $C_B$ respectively.

If one joins $C_A$ and $C_B$ by an edge $e$, then one constructs a complex $K$ satisfying $\pi_1(K) \simeq A \ast B$. Let $v_A$ and $v_B$ be the two end points of $e$ in $C_A$ and $C_B$ respectively. Let $r$ and $s$ the number of pre-images of $v_A$ and $v_B$ respectively in $\widehat{C_A}$ and $\widehat{C_B}$.

Let $C$ be the disjoint union of $s$ copies of $\widehat{A}$ and $r$ copies of $\widehat{B}$. Now, with respect to the natural projection $C \to K$, the points $v_A$ and $v_B$ have each one $rs$ pre-images. Then it is possible to construct a complex $\widehat{K}$ from the complex $C$ by joining each pre-image of $v_A$ by an edge to a pre-image of $v_B$, and conversely, in such a way that $\widehat{K}$ be connected.

Then $\widehat{K}$ is a finite covering of $K$ and $\pi_1(\widehat{K})$ is free, so the image of $\pi_1(\widehat{K})$ in $\pi_1(K) \simeq A \ast B$ is a finite-index free subgroup.

[In fact, the argument is closely related to Bass-Serre theory if we know graphs of spaces.]

Solution 4:

In fact, in his book (property 11 p. 120), Serre shows that the fundamental group of a finite graph of groups whose vertex groups are finite is virtually free. (The converse is also true, giving a characterization of virtually free groups.) I describe below the argument for the free product of two finite groups.

First, $A \ast B$ acts on a bipartite tree $T$ such that the edge stabilizers are trivial and the vertex stabilizers are conjugated to $A$ or $B$.

Let $|A|=n$, $|B|=m$ and $p=\max(n,m)$. Notice that $A$ (resp. $B$) acts freely on $A$ (resp. $B$) by multiplication on the left, hence a monomorphism $\varphi : A \hookrightarrow S_n \subset S_p$ (resp. $\phi : B \hookrightarrow S_m \subset S_p$). Now define $$ \psi = \varphi \ast \phi : A \ast B \to S_p,$$ and let $H= \mathrm{ker}(\psi)$. Clearly, $H$ is a finite-index subgroup of $A \ast B$. By construction, $H$ meets trivially $A$ and $B$; moreover, because $H$ is a normal subgroup, it meets trivially any conjugate of $A$ and $B$. Therefore, the action $H \curvearrowright T$, induced by $A \ast B \curvearrowright T$, is free.

We deduce that $H$ is a finite-index free subgroup of $A \ast B$; so $A \ast B$ is virtually free.