1. ## abstract algebra

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

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

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

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

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

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