Just trying to get my head around the proper way to understand the index of subgroups in infinite groups. Is the following claim true? If so, how would one prove it, and if not, can you help me come up with a counter example? And are there tighter restrictions (such as normality?) that make the claim true in this case?

Suppose that $G$ is a group (not necessarily finite), with subgroups $H$, and $K$ with $H \subseteq K \subseteq G$ and $\left| G: K \right| = \ell = \left| G: H \right|$ for some $\ell \in \mathbb{N}_{ > 1}$. Then $H = K$.

Edit (to add some context, and disclaimer, some of what I say below may be inaccurate since I'm just studying this)

I am currently studying quotients of free groups, and their finite index subgroups, and in particular those that are kernels of surjective groups homomorphisms onto another group.

That is if $G$ is a quotient of a free group on $n$ generators, by $m$ relations, so that $G \cong \left< x_{1}, \dots , x_{n} \mid r_{1}, \dots , r_{m} \right> $, and $\phi : G \rightarrow S$ is a surjective group homomorphism onto a finite group $S$ of size $s$ say. Then for $K = \operatorname{ker}(\phi)$, we see that $K$ is an index $s$ normal subgroup of $G$.

Then, $G$ can be realised as the fundamental group of a finite cell complex with $1$ $0$-cell, (labelled $b$) $n$ $1$-cells (labelled $x_{1}$ through $x_{n}$), and $m$ $2$-cells (attached along $r_{1}$ through $r_{m}$ respectively). This can then be triangulated to realise $G$ as the fundamental group of a finite path-connected simplicial complex $M$ say. Then, since $K$ is normal in $G$, it corresponds uniquely to a regular based covering map $p : (\tilde{M},\tilde{b}) \rightarrow (M, b)$ such that $p_{*}(\pi_{1}(\tilde{M})) = K$ (where $p_{*}$ denotes the group homomorphism induced by $p$).

Then, since $M$ is a finite path-connected simplicial complex, $\tilde{M}$ is too, and additionally $p$ is a simplicial map. Then, we can calculate a finite generating set for $p_{*}(\pi_{1}(\tilde{M}))$ by calculating the maximal tree $T$ in $\tilde{M}$ and then there is a generator for each edge in $\tilde{M}$ that is not in $T$. Then the relations come from the loops transversed in $\tilde{M}$ corresponding to each $r_{i}$ starting at each vertex in $p^{-1}(b)$. Then we know that the degree of $p$ corresponds to the index of $K$ in $G$, so $\tilde{M}$ has $s$ vertices. Then, since $p$ is regular and $M$ has $2n$ edges coming out of its singular vertex, $\tilde{M}$ has $2ns$ edges counting each edge twice, and so $\tilde{M}$ has $ns$ edges. Then $T$ has $s-1$ edges since it is a maximal tree of a graph with $s$ vertices. So we conclude that $K$ has a presentation with exactly $ns - (s-1) = (n-1)s + 1$ generators and $ms$ relations (we have $m$ relations for each point in $p^{-1}(b)$).

The motivation for this questions comes from the desire to be able to verify my solution. Since it is relatively easy to check if all of my generators are in the kernel, I can show containment easily. Then if I calculate indexes and show they are the same, would I be able to conclude that my solution is correct?

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    $\begingroup$ Notice how I've included the $=$ in the MathJax and only used the $s either side of groups of symbols. $\endgroup$ – Shaun May 25 '18 at 19:01
  • $\begingroup$ Thanks! Will make sure to do that in future. $\endgroup$ – Adam Higgins May 25 '18 at 19:04
  • $\begingroup$ You're welcome! Also, you'll find that simple "Here's the statement of my exercise, solve it for me" posts will be poorly received. What is better is for you to add context: What you understand about the problem, what you've tried so far, etc. Something to both show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$ – Shaun May 25 '18 at 19:07
  • $\begingroup$ @Shaun This isn't an exercise that I've been set. It's just something that I released I didn't know how to solve. The exact motivation for me realising I didn't know this fact is several layers deep in an otherwise unrelated problem that I'm working on (again, not an exercise I've been set). But I take your point, and will make sure to keep those tips in mind :) $\endgroup$ – Adam Higgins May 25 '18 at 19:09

Yes, if $H \le K \le G$ then $|G:H| = |G:K||K:H|$, so if $|G:H| = |G:K|$ is finite, then $|H:K|=1$ so $H=K$.

Proof: Let $T$ and $U$ be right transversals of $H$ in $K$ and of $K$ in $G$, respectively. Then $TU = \{tu : t \in T, u \in U \}$ is a right transversal of $H$ in $G$.


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