# subgroup question

#### 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?

#### tonio

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

• wutang