Third Friday: Free Subgroups and Sunshine

Today’s lecture was held outside, in the lovely warmth and sunlight. We considered two ways to approach the following statement.

Claim. If $H \leq F_n$ , then $H$ is free.

A Geometric Group Theory Proof

Instead of proving this claim directly, we started with a more general statement.

Theorem. If $G \curvearrowright X$ freely and without edge inversions, then $G$ is free.

Proof. Let $T$ be a tree, and consider an action $G \curvearrowright T$ such that the action is free and without edge inversions. Let $\mathcal{F}$ be a fundamental domain of $T$ over this action.

If $\mathcal{F} \cap g \mathcal{F} \neq \emptyset$ , then this intersection must consist exactly of points $v$ such that $v$ is the midpoint of an edge. This follows from how we construct fundamental domains and the definition of a tree.

In particular, since trees have no cycles, the fundamental domain of a tree must consist of endpoints, edges, and half-edges. If endpoints or edges were shared between two fundamental domains, then $\mathcal{F}$ would no longer be a partition $T$ . However, if $\mathcal{F}$ contains a half-edge, then the other half of the half-edge must belong to some multiple $g\mathcal{F}$ , again since we’re partitioning all of $T$ . If the midpoint of the edge belonged to one or neither, we wouldn’t be able to multiply any $h\mathcal{F}$ to obtain the midpoint, so the midpoint $v$ must belong to exactly two fundamental domains.

We want to use these midpoints to prove that $G$ is acting freely. To that end, we’ll seek to show that $S$ generates a (free!) basis for $G$ . Symbolically, we want to show that for

$S = \{g \in G \ | \ \mathcal{F} \cap g\mathcal{F} \neq \emptyset\},$

that $\langle S \ | \ \rangle = G$ .

Since $G$ acts freely on $T$ , we expect any for $e \neq t \in T$ that $g \cdot t \neq t$ , and by definition of a fundamental domain for an action, we expect for $t \in \mathcal{F}$ and $g \neq e$ that $t \notin g\mathcal{F}$ . Therefore every $g \in G$ satisfies $\mathcal{F} \cap g\mathcal{F} \neq \emptyset$ , and so $S$ contains all of $G$ .

We additionally seek to show that no freely-reduced non-trivial word $s = s_1 \ldots s_n \in S$ is trivial in $T$ . Toward that goal, consider some midpoint on our tree, an arbitrary $v_0 \in V(T)$ . We may consider the path generated by “building up” to $s$ , or more precisely, the sequence of vertices generated by

$s_n v = v_1$

$s_{n-1}s_n v = v_2$

$\ldots$

$sv = v_n$

With these midpoints on our tree in mind, we can consider the unique reduced path generated by connecting $v_0$ to $v_1$ , $v_1$ to $v_2$ , and so on. We’ll name these paths $\gamma_1, \gamma_2$ , and so on. Trees have pretty rigid structure; we can take advantage of this structure in order to make some pretty rigorous statements about each $\gamma_i$ . Toward a contradiction, suppose that we have a loop, or more precisely that the same $v_i$ occurs more than once in the path from $v_1$ to $v_n$ . Since $T$ is a tree and thus contains no cycles, the loop implies we must have backtracked:

But we see that a backtrack implies that three separate fundamental domains all shared the same midpoint edge; this contradicts the definition of an edge midpoint. Instead, it must be the case that every non-trivial word in $S$ corresponds to a non-trivial action. Thus we know that freely-reduced non-trivial words in $S$ are non-trivial on $T$ , and so $S$ is indeed a basis of $G$ , and since our basis includes no relators, $G$ is free. $\square$

Our original claim, that subgroups of free groups are themselves free, follows as a corollary. In particular, if $G$ is free and $H \leq G$ , then $G$ and $H$ both act freely on the Cayley graph of $G$ . Since $H$ acts freely, $H$ is free. $\square$ .

A Topological Proof

For the second part of the lecture, we considered a topological proof of our original claim that subgroups of free groups are free.

Definition. Given a space $X$ and a point $p \in X$ , a fundamental group $\pi_1 (X, p)$ is the group formed by the equivalence class of loops that start at $p$ over the operation of concatenation.

Example. Consider the torus $\mathbb{T}$ , and some point $p \in \mathbb{T}$ . We have (at least) two loops, one given by the vertical circle in red, and one given by the horizontal circle in blue:

Fact. For any graph $\Gamma$ , it’s the case that $\pi_1(\Gamma)$ is free.

I have very little background in topology, but I think this relates to the idea of thinking of free groups as little “bouquets” of loops. In some sense, I suspect this is what it means to be a free group.

Definition. Let $X$ and $Y$ be spaces, with a map

$f: Y \to X$

such that $f$ is continuous, surjective, and locally homeomorphic. Then $Y$ covers, or is a cover of, the space $X$ .

Fact. For some spaces $X, Y$ where $Y$ is a cover of $X$ , $\pi_1(Y)$ satisfies $Y \trianglelefteq X$ , and every $H \leq Y$ corresponds to an open cover of $X$ . This property is known as Galois Correspondence.

We can use this assortment of facts and definitions to show that subgroups of free groups are free. For some $\Gamma$ , it’s the case that $F_n = \pi_1(\Gamma)$ . By Galois Correspondence, we know that any $H \leq F_n$ implies $H$ is a cover of $\Gamma$ , and so $\pi_1(\Gamma)$ must be free.

2 Responses to Third Friday: Free Subgroups and Sunshine

Akash Ganguly says:

October 6, 2022 at 8:32 am

Great post John! I liked the little ending bit on fundamental groups — I really like how (at least for some more tangible objects) you can almost compute the fundamental groups by looking at it. For example, if you take a disc, you can smush (I think the actual term is homotope) any loop into a point. However, my intuition immediately fails when looking at higher homotopy groups. Is there any point in trying to visual these? Or maybe we should just look at them algebraically?

- John Eichelberger says:
  
  October 14, 2022 at 1:56 am
  
  Thanks! I can’t speak much to homotopy groups — I don’t have much experience using them. My sense is that past three dimensions, you can probably build an intuition for how these behave, but I don’t have that intuition and wouldn’t trust any result I couldn’t prove algebraically. I think about the way you can cheat with free groups and crunch everything into two dimensions; I wonder how true that is for more general spaces.

2 Responses to Third Friday: Free Subgroups and Sunshine

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