Showing that $ϕ(x)=x^n$ is a homomorphism from $G\to Z(G)$

Question

Let $G$ be a group with $|G:Z(G)|=n$ then $\phi(x)=x^n$ is a homomorphism from $G$ to $Z(G)$.

I guess it has a proof using transfer theory, I wonder whether it has an elemantary proof or not. Thanks.

Answer 1

Let $G$ be a group such that $Z(G)$ is of finite index $n$ in $G$. Then, the map $G \to G$, $g \mapsto g^n$ is a homomorphism.

Direct proof. Choose a system of representatives $R \subseteq G$ of $G/Z(G)$, $n=|R|$. The natural $G$-action on $G/Z(G)$ induces a $G$-action $e : G \times R \to R$ such that $gr Z(G) = e(g,r) Z(G)$, i.e. $e(g,r)^{-1} gr \in Z(G)$. It suffices to prove that $$ \prod_{r \in R} (e(g,r)^{-1} gr) = g^n. ~~~ (\star)$$ In fact, for $g,h \in G$ this implies $$(gh)^n \stackrel{(\star)}{=} \prod_{r \in R} e(gh,r)^{-1} ghr = \prod_{r \in R} e(g,e(h,r))^{-1} ghr = \prod_{r \in R} e(g,e(h,r))^{-1} g e(h,r) \cdot e(h,r)^{-1} hr\\ = \prod_{r \in R} e(g,e(h,r))^{-1} g e(h,r) \cdot \prod_{r \in R} e(h,r)^{-1} hr = \prod_{s \in R} e(g,s)^{-1} g s \cdot \prod_{r \in R} e(h,r)^{-1} hr\stackrel{(\star),(\star)}{=} g^n h^n.$$ To prove $(\star)$, we decompose the permutation $e(g,-) : R \to R$ into cycles. If $(r_1 \cdots r_k)$ is a cycle, i.e. $e(g,r_i)=r_{i+1}$ for $i<k$ and $e(g,r_k)=r_1$, we have $$\prod_{i=1}^{k} (e(g,r_i)^{-1} g r_i) = (e(g,r_k)^{-1} g r_k) \dotsc (e(g,r_2)^{-1} g r_2) (e(g,r_1)^{-1} g r_1) = r_1^{-1} g^k r_1$$ Since this lies in $Z(G)$, we also have $g^k \in Z(G)$. But then $r_1^{-1} g^k r_1 = g^k$. Since $\prod_{r \in R} (e(g,r)^{-1} gr)$ is the product of all these products, indexed by the cycles, and the cycle lengths add up to $n$, we get $g^n$. $\square$

Notice that this is basically the usual proof which uses the transfer map, but we only have to construct it in the special case and don't need to show that it's independent from the system of representatives.