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Suppose $H$,$K$ are normal subgroups of a group $G$. We don't assume $G$ is finite. Furthermore assume $[G:H]$ and $[G:K]$ are both finite. Then it follows that $[G: H \cap K]$ is divisible by $[G:H]$. Why is this?

Here's what I thought. Suppose we write $G = \bigcup_{i \in I} a_{i}(H \cap K)$ as a disjoint union of left cosets in $H \cap K$, here $I$ is some index set.

Now $a_{i}(H \cap K) \subset a_{i}H$ hence $G \subset \bigcup_{i \in I} a_{i}H$. So we get that $I$ is at least as big as $[G:H]$, thus $[G: H \cap K]$ divides $[G: H]$ no. ? Why is this wrong? why is the other way around?

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If you are just looking for an example to test your intuition, try $G = \mathbb{Z}$. What is the index of $2\mathbb{Z} \cap 3\mathbb{Z}$? –  Robert Bell Mar 10 '11 at 18:22
    
@Robert Bell: It is $6$. My question is, how do we show rigorously the statement? –  user8079 Mar 10 '11 at 18:25
    
@Robert Bell: Nevermind the statement is easy to show, just figured it out. Thanks though. –  user8079 Mar 10 '11 at 18:40
    
possible duplicate of Subgroups and finite index –  Arturo Magidin Mar 10 '11 at 19:18
    
See math.stackexchange.com/questions/7002/… for a proof that if $H\subseteq K\subseteq G$, then $[G:H]=[G:K][K:H]$ (in the sense of cardinalities). The divisibility requirement follows in the finite index cases. –  Arturo Magidin Mar 10 '11 at 19:19

2 Answers 2

I assume that you are familiar with this result: if M,N,G are groups and N < M < G (where < means "to be a subgroup of"), then [G:N]=[G:M][M:N]. In your case, take N=H\cap K, and M=H and apply the result I mentioned.

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Your argument goes wrong here: "So we get that $I$ is at least as big as $[G:H]$, thus $[G: H \cap K]$ divides $[G: H]$". Since $|I|=[G: H \cap K]$, "$I$ is at least as big as $[G:H]$" means $[G: H \cap K] \ge [G: H]$, which doesn't go well with, let alone implies, "$[G: H \cap K]$ divides $[G: H]$".

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