1. ## abstract algebra

Prove that if $H$ is a normal subgroup of $G$ of prime index $p$ then for all subgroups $K$ of $G$ either
(i) $K$ is a subgroup of $H$, or
(ii) $G = HK$ and $|K : K$ intersect $H| = p$

I tried to do this in 2 cases. The first case is to let $K$ be a subest of $H$, then since $K$ is a subgroup of $G$, it is a subgroup of $H$. For case 2, I let $K$ not contained in $H$, then try to show $G = HK$ and $|K : K$ intersect $H| = p$, but don't know how... please help, thank you.

2. Let $\pi:G\to G/H$ be the quotient map and consider $\pi(K)$. (It is a subgroup of the simple group G/H.)

3. Originally Posted by Opalg
Let $\pi:G\to G/H$ be the quotient map and consider $\pi(K)$. (It is a subgroup of the simple group G/H.)
I don't get it...

4. Originally Posted by dori1123
I don't get it...
What Opalg means is this:

The group $G/H$ has prime order, therefore, it only has no proper non-trivial subgroup. The natural projection $\pi : G\to G/H$ defined by $\pi (a) = aH$ is a homomorphism. Thus, by the property of homomorphism, if $K$ is a subgroup of $G$ then $\pi (K)$ is a subgroup of $G/H$. This means that $\pi (K) = G/H$ or $\pi (K) = \{ H \}$. We have two cases, (i) $\pi (K) = G/H$ (ii) $\pi (K) = \{ H\}$. Now consider each case.

5. Originally Posted by ThePerfectHacker
if $K$ is a subgroup of $G$ then $\pi (K)$ is a subgroup of $G/H$. This means that $\pi (K) = G/H$ or $\pi (K) = \{ H \}$. We have two cases, (i) $\pi (K) = G/H$ (ii) $\pi (K) = \{ H\}$. Now consider each case.
How do you know $\pi (K) = G/H$ or $\pi (K) = \{ H \}$?

6. Originally Posted by dori1123
How do you know $\pi (K) = G/H$ or $\pi (K) = \{ H \}$?
Because $G/H$ has only non-proper and trivial subgroups. That is $G/H$ - the full group itself. And $\{ H \}$ - the set containing the identity element.