If G is a group of order p^n and N is a normal subgroup of order p where p is prime and n is natural number, then prove that N is contained in the center of G.
Asked
Active
Viewed 119 times
0
-
I did the following: Take any n in N. Then gng^-1 should be in N. Then there exists an m in N such that gn=mg. Also, we have order of n divides p. Then what should I do ? – Richard Nov 10 '13 at 07:57
1 Answers
0
I'll be a bit brief, but feel free to ask for clarification on any part of the proof that is unclear.
Just note that since N is normal, it must be a union of conjugacy classes. Each conjugacy class of G though must have $p^i$ elements for some $i \in \{0,\dots,n\}$. N is a subgroup so it must contain the identity, which is a conjugacy class itself. But we are given that it has order $p$, so the remaining elements must be their own conjugacy classes. To see this, note that if $N$ contained an element which belongs to a conjugacy class of order greater than 1, it would contain a conjugacy class of at least order $p$, so then we would have that $|N| \geq p+1$ which is clearly a contradiction.
Any element which is the only element in a conjugacy class must be in the center, so in particular, every element of $N$ is in the center of $G$.