Why doesn't the infinite dihedral group contain a free subgroup of rank 2?

Question

Our professor just told us that $D_{\infty}$ is "too small" whatever that means. Can someone prove this statement and give some reason as to why this statement holds true but doesn't for other free products $A\star B \cong G$ and $G$ not isomorphic to $D_{\infty}$.

Answer 1

I'll start by proving the results you care about, and then talk about the word "small".

Lemma 1. $D_{\infty}$ contains an infinite cyclic subgroup of finite index.

Proof. Starting with the presentation of $D_{\infty}\cong C_2\ast C_2$ as a free product, we can use Tietze transformations to show that it has an infinite cyclic subgroup of finite index as follows: $$\begin{align*} C_2\ast C_2&=\langle a, b\mid a^2=1, b^2=1\rangle\\ &\cong \langle a, b, c\mid a^2=1, b^2=1, ab=c\rangle &\text{add new generator $c=ab$}\\ &\cong \langle a, b, c\mid a^2=1, b^2=1, b=a^{-1}c\rangle\\ &\cong \langle a, b, c\mid a^2=1, (a^{-1}c)^2=1, b=a^{-1}c\rangle\\ &\cong \langle a, c\mid a^2=1, (a^{-1}c)^2=1\rangle&\text{remove old generator $b$}\\ &\cong \langle a, c\mid a^2=1, a^{-1}ca^{-1}=c^{-1}\rangle\\ &\cong \langle a, c\mid a^2=1, a^{-1}ca=c^{-1}\rangle&\text{as $a^{-1}=a$}\\ &\cong\mathbb{Z}\rtimes C_2 \end{align*} $$ As $\mathbb{Z}\rtimes C_2$ contains an infinite cyclic subgroup of index $2$, the claim follows. QED

Lemma 2. If $G$ contains a cyclic subgroup of finite index, then so do all its subgroups.

Proof. Suppose $C$ is a finite index cyclic subgroup of $G$. Without loss of generality, we may assume that $C$ is normal in $G$ (for example, replace $C$ with the intersection of all the conjugates of $C$; this still has finite index in $G$ and is still cyclic). Let $H$ be an arbitrary subgroup of $G$.

Now, $HC/C$ is finite as it is a subgroup of $G/C$. From our isomorphism theorems, we have $HC/C\cong H/(H\cap C)$, and so $H\cap C$ has finite index in $H$. As $H\cap C\leq C$ is cyclic, the result follows. QED

Theorem. $D_{\infty}$ does not contain a non-abelian free group.

Proof. By Lemmas 1 and 2, every subgroup of $D_{\infty}$ contains an infinite cyclic subgroup of finite index. As non-abelian free groups do not have such subgroups (by the universal property of free groups, and as there are $2$-generated groups which do not have such subgroups), the result follows. QED

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The second result you want to understand is that every other free product $A\ast B\not\cong D_{\infty}$, and $A,B$ non-trivial, contains a free group of rank $2$. The proof of this depends on how you have "encountered" free products.

If you encountered them combinatorially, as the group consisting of words over the elements of $A$ and $B$ then take non-trivial elements $a\in A$, $b\in B$ and consider the subgroup $\langle ab, ba\rangle$. If neither $a$ nor $b$ has order $2$, then every non-trivial word $W$ over $(ab)^{\pm1}, (ba)^{\pm1}$ is non-empty after reduction, and the result follows. If precisely one of them has order $2$, then the same is true but more thinking is required. If all the non-trivial elements of $A$ and of $B$ have order $2$, then you need to take a different generating pair but the same idea works.

If you encountered free products via their actions on trees, then you know that $A*B$ acts properly on a tree $T$ which is not a line. Now take two elements $x, y\in A\ast B$ such that:

1. neither element fixes a point, and
2. they have no "common axis" - each of them can be associated to a line, or axis, in $T$ which they act along by translation, and as $T$ is not a line we can pick $x$ and $y$ such that these axes are different. Then $\langle x, y\rangle$ acts freely on a subtree of $T$, and hence is free.

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Lets now discuss "too small", in reference to $D_{\infty}$. In the world of finitely presented groups, it is helpful to understand how different classes of groups are connected to one another. There are a number of ways of doing this, but the overwhelming preference since at least the 1980s is for connections which are stable up to finite index ("commensurability invariant"). Therefore, $D_{\infty}$ is essentially equivalent to $\mathbb{Z}$, which is a "small" group.

Viewpoint 1: Bridson’s Universe of Finitely Presented Groups. In an article accompanying his 2006 ICM lecture, Martin Bridson drew what he considered to be the "Universe of Finitely Presented Groups". This website gives a reasonable explanation and a link to the article, but here is the Universe:

There are a lot of question marks, but the main point I want to communicate is that the left "dot" on the cone is labelled $\mathbb{Z}$ and corresponds to all virtually-$\mathbb{Z}$ groups; everything else is more complicated. So "small" seems a reasonable description here!

Viewpoint 2: The Largeness ordering. In the late 1970s/early 1980s, Steve Pride defined an ordering $\leq_L$ on the class of finitely presentable groups:

$A\leq_LB$ if $B$ contains a subgroup of finite index which surjects onto $A$.

Pride then called a group large if is larger than every other finitely presented groups; non-abelian free groups are large (they contain finite index free subgroups of arbitrarily large rank), as are group which admit a presentation $\langle\mathbf{x}\mid\mathbf{r}\rangle$ with $|\mathbf{x}|\geq |\mathbf{r}|+2$ (this is a theorem of B. Baumslag and Pride).

Virtually-$\mathbb{Z}$ groups sit at the bottom of this ordering - they are only "larger" than other virtually cyclic groups. Therefore, with this ordering in mind, it is reasonable to call them "small".