- #1
Omukara
- 9
- 0
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 -__-' :)
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 -__-' :)