# Show $\vert G \vert = \vert HK \vert$ given that $H \trianglelefteq G$, $G$ finite and $K \leq G$.

My problem is a variation of one in Dummit and Foote:

Let $G$ be a group and $H \trianglelefteq G$. Prove: If $G$ is finite and $[G:H] = p$, a prime number, then for any $K \leq G$, either $K \leq H$ or $G = HK$ and $[K : H \cap K] = p$.

Ok, so I suppose $K \not\leq H$. I am trying to show $\vert G \vert = \vert HK \vert$. The second part follows directly from that and using Lagrange's Theorem in a nice way.

I just need a hint on how to show $\vert G \vert = \vert HK \vert$. I've been using Lagrange's theorem and some corollaries so far, but haven't found the right approach. I'm not allowed to use Isomorphism theorems or quotient groups on this problem. Professor hinted it should follow directly from lagrange since we suppose $G$ is finite.

Any SMALL hints would be greatly appreciated!!!

Edit: I'm using that $\vert HK \vert = \frac{\vert H\vert \vert K \vert}{\vert H \cap K \vert}$ a lot too.

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HINT: Use the normality of $H$ to show that $HK$ is a subgroup of $G$ (that clearly contains $H$).

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I already know that $HK$ is a subgroup of $G$ and $H, K \leq HK$ and $H \cap K \leq H$ and $H \cap K \leq K$. So I'm trying to use Lagrange and these propositions to show that $\vert G \vert = \vert HK \vert$ showing that $G = HK$. –  29193 Sep 28 '12 at 21:22
@Robert: Then you’re really done: $H\le HK\le G$, so $$|G|=[G:HK]\cdot|HK|=[G:HK]\cdot[HK:H]\cdot|H|\;.$$ –  Brian M. Scott Sep 28 '12 at 21:30
I had that written down a few pages ago, but didn't see what it said about $\vert HK \vert$, so then I started solving for $\vert HK\vert$ and trying to use other formulas and didn't really get anywhere. I specifically tried to break it down into $\vert G \vert = \vert HK \vert$. –  29193 Sep 28 '12 at 21:34
@Robert: Since $|G|=[G:H]\cdot|H|$, it gives you $$p=[G:H]=[G:HK]\cdot[HK:H]\;.$$ –  Brian M. Scott Sep 28 '12 at 21:35
@Robert: But look at it: it expresses $p$ as a product of two integers! –  Brian M. Scott Sep 28 '12 at 21:45

Since $H$ is normal in $G$, If $K \le G$, then $HK \le G$. By Lagrange's theorem $|HK|$ divides $|G|$. So $|HK| = \frac{|H||K|}{|H \cap K|}$ divides $|G|$. So $\frac{|K|}{|H \cap K|}$ divides $[G : H] = p$. So $\frac{|K|}{|H \cap K|}$ is either $1$ or $p$. If $\frac{|K|}{|H \cap K|} = 1$, we easily see that $K \le H$. In the other possibility $\frac{|K|}{|H \cap K|}= p$. So $|HK| = p \cdot |H| = |G|$. So $G = HK$.

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Think about whether $H$ and $K$ have to be disjoint or whether they can overlap in $G$. In the case that they are disjoint, then $\lvert H\cap K\rvert = 1$.

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