1. ## subgroup question

Show that if G is a group of order 168 that has a normal subgroup of order 4, then G has a normal subgroup of order 28.
Here is what I have so far:
Let H be a normal subgroup of order 4. Then |G/H|=42=2*3*7, so then G?N has a unique, and therefore normal Sylow 7-subgroup, lets call it K.
I was told to use the correspondence theorem, but I am not sure where it works in here. any ideas?

2. Originally Posted by wutang
Show that if G is a group of order 168 that has a normal subgroup of order 4, then G has a normal subgroup of order 28.
Here is what I have so far:
Let H be a normal subgroup of order 4. Then |G/H|=42=2*3*7, so then G?N has a unique, and therefore normal Sylow 7-subgroup, lets call it K.
I was told to use the correspondence theorem, but I am not sure where it works in here. any ideas?

So $\displaystyle G/H$ has a unique, normal sugroup $\displaystyle K/H$ of order 7, which then pulls back under the canonical projection (this is the correspondence theorem) to

a subgroup $\displaystyle K\leq G$ of order $\displaystyle 7\cdot 4=28$ (why?!? Look at the respective indexes which, again by the CT, stay the same...) which, again by the corr. theorem, is

also normal in $\displaystyle G$

Tonio