Intersection of two groups of finite index is a group of finite index

Question

These questions may occur to be simple, but I can't wrap my mind around them.

Why the intersection of two groups of finite index is a group of finite index?
Why if $g\in G$ has finitely many conjugates then the centralizer $C_G(g)$ has a finite index in $G$?

I have these questions when try to understand the answer to this post.

It might help for you to relate the index of a subgroup with the cardinality of the set of equivalence classes obtained after quotienting the parent group by the subgroup. — Sinister Cutlass, Nov 12 '15 at 21:53
@SinisterCutlass, this relation I have in mind, since this is the definition of index =) — Glinka, Nov 12 '15 at 21:56

score 4 · Accepted Answer · answered Nov 12 '15 at 22:21

4

Let $G$ be a group and $H, K$ be subgroups. $H, K$ have finite index iff the $G$-sets $G/H, G/K$ are finite. Now I claim that $G/(H \cap K)$ is naturally isomorphic to the orbit containing $(e, e)$ in the product $G$-set $G/H \times G/K$, which is necessarily finite (and in fact has size at most $|G/H| \times |G/K|$).
The conjugates of $g \in G$ are the orbit containing $g$ in the $G$-set given by $G$ itself with the conjugation action. The centralizer $C_G(g)$ is the stabilizer of $g$ with respect to this action. Hence this orbit is $G/C_G(g)$ and the index of $C_G(g)$ in $G$ is precisely the number of conjugates of $g$.

answered Nov 12 '15 at 22:21

Qiaochu Yuan

What do you mean by $G$-set? – Glinka Nov 12 '15 at 22:24
@Glinka: I mean a set equipped with an action of $G$. – Qiaochu Yuan Nov 12 '15 at 22:25
Is it possible to show it without using terminology of the group action? If it is, should ask another question for that? – Glinka Nov 12 '15 at 22:29
1

@Glinka: it's going to be pretty hard to do any group theory if you don't know about group actions... – Qiaochu Yuan Nov 12 '15 at 22:34
Ok, I'm seeing into this action thing. I get now what you did in (2). In (1) what action do we consider? How $G$ acts on $G/H$, $G/K$ and so on? – Glinka Nov 12 '15 at 23:55
@Glinka: $G$ acts diagonally: if $x \in G/H, y \in G/K$, then $g(x, y) = (gx, gy)$. Using this it's not hard to compute that the stabilizer of $([e], [e])$ is $H \cap K$. – Qiaochu Yuan Nov 13 '15 at 01:23

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