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?