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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.

3 Answers 3

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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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    @Robert: Now you’ve got it! (We’ve all had those momentary blind spots from time to time.)2012-09-28
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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 $\lvert H\cap K\rvert = 1$.