Sylow's Theorem (normal subgroups)

[Omukara]

Hello,

I'm having difficulty understanding how to solve the following question using Sylow's Theorem:

Suppose G is a group of order 56. Show that either:
i) G has a normal subgroup of order 7 or
ii) G has a normal subgroup of order 8.


I started by decomposing 56 into desired form (kp^n): 56 = 2^3*7,

My lecturer insisted that when we do Sylow counting to take the larger prime as p, so I initially take p=7(?)

So by Sylow's theorem there are N=1 or N>=8 subgroups of order 7.

I get confused at this point. Is it possible that we have a normal subgroup for each of the cases of N (i.e. when N=1 and N=8)? And then why does this not allow for normal subgroups of order 8?


Help will be tremendously appreciated! Algebra exam in a few days -__-' :)

 

[micromass]

Hi Omukara!

A Sylow subgroup is normal if and only if there is only one Sylow subgroup of that order.
So, in your example, take the Sylow subgroups of order 7. You have correctly deduced that there are two cases. There are either 1 Sylow subgroup of order 7 and there are either 8 Sylow subgroups of order 7.

Now, in the former case, the unique Sylow subgroup of orfer 7 is normal. In the latter case, there are 8 Sylow subgroups, which are not normal. So it suffices there to show that there can only be one subgroup of order 8 (if there is only one, then it must be normal!).

So, how do we do this? We count the number of elements of order 7. You know that there are 8 subgroups of order 7, so, how many elements of order 7 are there then?

 

[Omukara]

ah, fantastic!:) Thanks a lot!

...so there are 8(7-1) = 48 elements of order 7, leaving 8 elements after 56 - 48.

And since we know there is a subgroup of order 8, we can now deduce that this is indeed unique.

Hence it follows; subgroup of order 8 is a normal subgroup in G.

Got it - thanks again!^_^