# Prove that intersection of finite index subgroups has finite index.

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.

• 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. – Akash Gaur 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. – Akash Gaur 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

$G$ 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.