A virtually cyclic group $G$ cannot contain $F_2$

Question

A virtually cyclic group $G$ cannot contain $F_2$.

This looks easy. Let $H = \langle a \rangle$ be a cyclic subgroup of $G$ with finite index. So let $G = g_1 H \sqcup g_2 H \sqcup \cdots \sqcup g_n H$, where $n \in \mathbb{N}$ and $g_1 = $ identity in $G$. Let $\phi: F_2 = \langle x,y | -\rangle \longrightarrow G$ be an injection. Then $\phi(x)$ and $\phi(y)$ are infinite order elements in $G$. Let $\phi(x) = g_i a^s$ and $\phi(y) = g_j a^t$. And now no matter what I do I cannot show that this cannot be possible. Maybe it is too simple but I have already spent a lot of time.

Answer 1

Let $\phi\colon F_2\to G$ be a morphism. We will prove it is not one-to-one.

Since $H$ is of finite index in $G$, it contains a normal subgroup $N$ of finite index in $G$. In particular, there exist positive integers $r$ and $s$ such that $\phi(x)^r=\phi(x^r)$ and $\phi(y)^s=\phi(y^s)$ lie in $N$. Since $N\subseteq H$, then $\phi(x^r)$ and $\phi(y^s)$ commute. Therefore, $\phi([x^r,y^s])=[\phi(x^r),\phi(y^s)]$ is trivial, proving that $\phi$ is not one-to-one, since in $F_2$ no nontrivial power of $x$ commutes with a nontrivial power of $y$.

Note that we don't need $H$ to be cyclic, just abelian.

Answer 2

Here is another proof. A cyclic group is abelian, hence amenable, and a virtually cyclic group is virtually amenable, hence amenable itself. But $F_2$ is not amenable, so it cannot be a subgroup of a virtually cyclic group. This uses the fact that closed subgroups of amenable groups are amenable. So we have also shown that a virtually abelian group cannot contain $F_2$.

Reference: These notes by Alain Valette.

Answer 3

Here is yet another proof.

Suppose $F \subset G$ is a subgroup of $G$ isomorphic to $F_2$.

Since the infinite cyclic subgroup $H$ has finite index in $G$, it follows that $F \cap H$ has finite index in $F$ and $F \cap H$ is infinite cyclic.

However, there is a nice formula which says that if $n \ge 2$ then an index $m$ subgroup of $F_n$ is isomorphic to the free group $F_{1 - m + nm}$ (you can prove this formula with a bit of topology, the same bit used to prove that a subgroup of a free group is free).

So an index $m$ subgroup of $F_2$ is isomorphic to $F_{1 - m + 2m} = F_{1 + m}$. But $m \ge 1$, so $1+m \ge 2$, and $F_{1+m}$ is not infinite cyclic.

Answer 4

This actually works in more generality, i mean:

if $G$ contains a copy of $F_2$, and $H$ is a finite index subgroup of $G$, then $H$ must also contain a copy of $F_2$ (this is an exercise in Clara Loh’s textbook).

For a proof, this is my way to do it:

Let $F := \langle x, y \rangle \subset G$ be freely generated by $x$ and $y$. Since $H$ has finite index in $G$, there must be some $g \in G$ such that $x^n, x^m \in gH$ where $n > m > 0$. So, there are $a, b \in H$ such that $x^n = ga$ and $x^m = gb$. Now, $x^mb^{-1} = g$ and therefore $x^{n - m} = b^{-1}a \in H$. Similarly, we find a power of $y$ living in $H$, and take the subgroup generated by them, which can easily be shown to be free of rank 2.

Now, back to the initial question, we just notice that a cyclic group can’t contain $F_2$.

Answer 5

And another proof. This works in the more general of $G$ being virtually solvable (which it is, if $G$ is virtually cyclic of course). This shows, for example that the two alternatives in the Tits alternative for linear groups (see here) are mutually exclusive.

Since $G$ is virtually solvable, we have that $H\subseteq G$ of finite index is solvable. As the other answers say, having a copy of $F_2$ in $G$ implies that we have a copy in the finite index subgroup $H$ as well.

So, now we have that $F_2$ is a subgroup of a solvable group. But subgroups of solvable groups are solvable. So $F_2$ must be solvable.

And quotients of solvable groups are also solvable. So, every quotient of $F_2$ (which is any two-generated group you want!) is solvable. This is clearly not true. For instance, $S_5$, the symmetric group on $5$ letters is generated by a transposition and a cycle, and is not solvable.