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I'm trying this problem from Herstein:

Q) If G is a group and H, K are two subgroups of finite index in G, prove that H $\cap$ K is of finite index in G. Can you find an upper bound for the index of H $\cap$ K in G?

My attempt:

$$\left [ G:H \right ]= \frac{|G|}{|H|} < \infty \wedge \left [ G:K \right ]= \frac{|G|}{|K|} < \infty $$ $$H\leq G \wedge K\leq G \rightarrow H\cap K\leq G$$ $$\left |H\cap K \right |=\frac{\left | H \right |\left | K \right |}{\left | HK \right |}\rightarrow\left [G:H\cap K \right ]=\frac{\left | G \right |\left | HK \right |}{\left | H \right |\left | K \right |}$$ $$HK\subseteq G\rightarrow \left |HK\right |\leq \left | G \right |\rightarrow\frac{\left |HK\right |}{|K|}\leq \frac{|G|}{|K|}< \infty$$ $$\rightarrow \left [G:H\cap K \right ]=\frac{\left | G \right |\left | HK \right |}{\left | H \right |\left | K \right |} \leq [G:H][G:K]<\infty$$

The problem seems to be in the 4th step as $|HK|,|K|$ and $|G|$ are all $\infty$.

I've found other solutions to this problem on MSE that I've understood. I just wanted to know if this approach had any merit.

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For a finite group, every subgroup will be finite index, and for an infinite group, methods that count the size of the subgroups are right out. A more refined approach will be necessary. – Aaron Aug 4 '14 at 5:16
@Aaron Thanks, what i meant was "finite order subgroups H,K". That seems necessary for step 4 to hold. – Rhaldryn Aug 4 '14 at 5:25
Now I'm confused, do you want finite order or finite index? In an infinite group, a subgroup can be one or the other (or neither), but never both. – Aaron Aug 4 '14 at 5:32
@Aaron You're right. I'm just trying too hard to find a case where the above method would work. I'll just go with the other solutions. – Rhaldryn Aug 4 '14 at 5:37
Note that in this solution you have used $|K|$ such as in $\frac{|G|}{|K|}<\infty$ but if $G=\mathbb{Z}$ and $K=2\mathbb{Z}$ then this is somewhat problematic. – Belgi Aug 4 '14 at 5:49

1 Answer 1

up vote 4 down vote accepted

$G$ acts on the set $G/H\times G/K$ by coordinatewise multiplication, i.e. it sends $g(xH,yK)\to (gxH,gyK)$. Consider now the element $(H,K)$. Then ${\rm stab}(H,K)=\{g\in G:gH=H,gK=K\}=H\cap K$. This means, by the orbit stabilizer theorem, that $|G:H\cap K|=|\mathcal O(H,K)|$, the orbit of $(H,K)$ under multiplication by $G$. But $H,K$ are of finite index, so this orbit has finitely many elements. In fact, we have shown that $|G:H\cap K|\leqslant |G:H||G:K|$ even in the non-finitary case.

You're writing $|G:H|=|G|/|H|$. This makes sense only if $G$ is finite, but the question, in such case, is easily answered. What is true, however, is that as cardinal numbers $|G:H||H|=|G|$. See here for a (correct) proof that salvages your argument.

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The OP clearly stated "I've found other solutions to this problem on MSE that I've understood. I just wanted to know if this approach had any merit." So, does Rhaldryn's approach have any merit? Is this answer related in any way to Rhaldryn's approach? That is, answer the question please... – user1729 Aug 4 '14 at 8:06
Well, fine, you should edit your answer to say this! Just spouting a proof of the result is not really relevant, especially as the OP has pointed out that they have seen and understood proofs of the result at MSE (so clearly your answer is a duplicate!). If you want to answer the question other than just saying "no, here are your problems" then you might consider suggesting how the problems can be resolved. Perhaps your answer does this already (I do not know, I didn't read it in detail), but if so then you should make this clear. – user1729 Aug 4 '14 at 8:18
Thanks, the top voted answer on the page you linked is what I was looking for. I haven't studied OST in detail so I wanted to avoid that. – Rhaldryn Aug 4 '14 at 11:27
@user1729 That's my bad, I made that edit later though I did initially mention that I'd prefer an extension of my argument if it was indeed correct. – Rhaldryn Aug 4 '14 at 11:30

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