subgroup question

Apr 2010
51
0
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?
 
Oct 2009
4,261
1,836
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
 
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