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

t.b.
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Balin
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    Wikipedia claims yes, without proof or source. – hmakholm left over Monica Jan 02 '12 at 01:40
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    My guess is that this follows from Bass-Serre theory (http://en.wikipedia.org/wiki/Bass%E2%80%93Serre_theory) but I can't supply any details. – Qiaochu Yuan Jan 02 '12 at 01:51
  • Take elements of the form $ab$ (with $a \in A\setminus{1}$ and $b \in B\setminsu {1}$. These will tend to generate free subgroups, I think. – Matt E Jan 02 '12 at 01:53
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    This follows very easily from the Kurosh subgroup theorem. As t.b. notes, the kernel of the map from $A\ast B$ to $A\times B$ is a normal subgroup of finite index, and since the map is injective on both $A$ and $B$, the Kurosh subgroup theorem guarantees the kernel is free. –  Jan 02 '12 at 07:30
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    If you do not want to use the Kurosh subgroup theorem to prove Steve D's comment, you can induct on the number of commutators you are multiplying together (on $k$, where $[g_1, h_1]^{\epsilon_1}\ldots [g_k, h_k]^{\epsilon_k}$ where no commutator appears before its inverse). You want to show that (via induction) such a product will always have length $\geq k+3$ and will end with $g_kh_k$ if $\epsilon_k=1$ or $h_kg_k$ if $\epsilon_k=-1$. Can you see why this would prove the result? – user1729 Jan 04 '12 at 13:29

4 Answers4

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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 No 1.1.3. (or page 6 of the English translation).

t.b.
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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.

Geoff Robinson
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  • I'm sorry but I do not understand the argument. I'm not sure what the precise general conditions are for $\chi(G) = \dfrac{1-r}{[G:F]}$ to hold, but doesn't this assume that you already know that there is a free subgroup of finite index? – t.b. Jan 02 '12 at 14:24
  • Yes, I did not explain it very well. The argument is really that if the Euler characteristics of groups $A$ and $B$ and $C = A \cap B$ exist, then the Euler characteristic Euler characteristic of $A*_C B$ exists and is equal to $\chi(A) + \chi(B) - \chi(C).$ The existence of the Euler characeristic implicitly indicates the existence of the free subgroup of finite index. The Euler characteristic for a finite group $X$ is always defined, and is equal to $\frac{1}{|X|}.$ The Euler characteristic of a group $Y$ with a free subgroup $F$ on $t$ generators of finite index is $\frac{1-t}{[Y:F]}.$ – Geoff Robinson Jan 02 '12 at 14:38
  • Ah, I see. Thank you for the explanation. – t.b. Jan 02 '12 at 14:44
  • Note that the Euler characteristic tells us that the free subgroup of index $|A|.|B|$ given in your answer is free on $(|A|-1)(|B|-1)$ generators. – Geoff Robinson Jan 02 '12 at 19:53
  • Yes, this is left a bit implicit in my answer, but I thought it should be clear: I identified the generating set explicitly in terms of the set of commutators $aba^{-1}b^{-1}$ with $a \in A \smallsetminus {1}$ and $B \smallsetminus {1}$ of which there are $(|A|-1)(|B|-1)$. Thanks for pointing it out. – t.b. Jan 02 '12 at 20:04
  • Yes, it was clear enough I suppose . I was just really pointing out that the Euler characteristic method gives a direct calculation of the of the rank of the free group. – Geoff Robinson Jan 02 '12 at 21:38
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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.]

Seirios
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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.

Seirios
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  • The proof in the comment of @user641 (and the answer of t.b.) are quite similar and give better lower bounds to the index of a finite rank free group, namely $nm$ as opposed to $\max(m,n)!$. – Lee Mosher Jun 03 '14 at 19:19