The \(p\)th power map in order \(p^3\) groups, where \(p\) is an odd prime (D&F 5.4.9)

Prove that if \(p\) is an odd prime and \(P\) is a group of order \(p^3\) then the \(p\)th power map is a homomorphism of \(P\) into \(Z(P)\). If \(P\) is not cyclic, show that the kernel of the \(p\)th power map has order \(p^2\) or \(p^3\). Is the squaring map a homomorphism in non-abelian groups of order 8? Where is the oddness of \(p\) needed in the above proof?

If \(P\) is abelian, then \(Z(P)\) has order \(p^3\), and the first statement immediately follows. Otherwise, then \(Z(P)\) can only have order \(p\), since \(Z(P)\) is nontrivial and \(G/Z(P)\) cannot have order \(p\). So the order \(p^2\) group \(G/Z(P)\) is abelian, implying \(P' \le Z(P)\). But \(P'\) is non-trivial since \(P\) is non-abelian, forcing equality. In particular, every element of \(P'\) has order \(p\). Now notice that if \(p\) is odd then \(\frac{p-1}{2}\) is an integer. Since \([y,x]^{\frac{p-1}{2}} \in P'\), we have \([y,x]^{\frac{p(p-1)}{2}} = ([y,x]^{\frac{p-1}{2}})^p = 1 \in Z(P)\). Thus \((xy)^p = x^p y^p\) (by 5.4.8), proving the map is a homomorphism. (Note that this conclusion fails if \(p = 2\), since \((xy)^2 = x^2 y^2 [y,x]\).)

The next step is to show that for non-abelian groups \(P\), \(x^p \in Z(P)\) for all \(x\). This is trivial if \(x\) has order \(p\), so suppose it has order \(p^2\). Then the subgroup \(\langle x \rangle\) is normal in \(P\) (having index \(p\)). If \(Z(P) \nleq \langle x \rangle\), then their intersection is trivial; this combined with the fact that both groups are normal implies \(P \cong Z(P) \times \langle x \rangle\), contradicting the fact that \(P\) is non-abelian. So \(Z(P)\) must be the unique subgroup \(\langle x^p \rangle\), and the conclusion follows.

If \(P\) is non-abelian, then the the image is a subgroup of \(Z(P)\) having order \(1\) or \(p\). If \(P\) is abelian and non-cyclic, then it is isomorphic to \(Z_{p^2} \times Z_p\) or \(E_3\). Letting \(Z_{p^2} = \langle x \rangle\) in the former case, its image is \(\langle x^p \rangle \times 1\) which has order \(p\). In the latter case, the image is trivial. Applying the First Isomorphism Theorem to all these cases, we can see that the kernel has order \(p^2\) or \(p^3\).