Prove that if p is an odd prime and P is a group of order then the power map is a homomorphism of P into Z(P). If P is not cyclic , show that the kernel of the map has order or . Is the squaring map a homomorphism in nonabelian groups of order 8? Where the oddness of p needed in the above proof?
Hint: Assume x, y in G and both x and y commute with . Prove that for all n in Z+ ,
My problem is I can't see why the hint holds.I aldo don't understand where we
use the hint in the original problem.Can anybody please help?
Wait... if , then shouldn't ?
Even then, of course, this would force to be abelian. There are only 2 (abelian) groups of order ; cyclic and elementary abelian. The group cannot be cyclic, though, else would be abelian. Thus
,
so we get what you wanted to conclude from that fact anyway.
I'm gonna give more of a hint, because I thought about this one myself and I think the way to proceed was not at all obvious (unless there's an easier way I missed).
Let's assume the hint. Note, then, that to prove that the -power map is a homomorphism, it suffices (by the hint) to prove that (the identity of ) for all .
Assume that is nonabelian (otherwise the result of the problem is trivial). First, we find the order of . All -groups have nontrivial centers, and is nonabelian; thus, . If , then is cyclic is abelian, obviously a contradiction.
So we conclude .
Then is abelian.
Recall the following properties of the commutator subgroup of a group :
- If such that the quotient group is abelian, then .
- if and only if is abelian.
Using this, can we say what the order of is? And what does this say about the equation in the hint, when ?
I believe we can use the Burnside Basis Theorem here (see p140 of Robinson's book, `A Course in the Theory of Groups'). It actually yields quite a neat proof, which is why I post this...it is also one of my favourite theorems!
Now, because we are in a non-abelian group of order we must have that , by the BBT.
Also by the BBT we have that .
Therefore, it is sufficient to prove that , which it is (see topspin's post).
I am, however, a bit confused as to where we need odd primes. Certainly this proof does not rely on odd primes, glancing over at p141 of Robinson I see that the identity in the hint doesn't need odd primes either, and I'm pretty sure topspin's proof doesn't need odd primes...also, the result does hold for the Quaternion group, and for , the two non-abelian groups of order 8...
EDIT: for notation, the Frattini subgroup .