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

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.

• Yes, $x \mapsto x^n$ is exactly the transfer homomorphism from $G$ to $Z(G)$. I am not sure what you mean by elementary. The transfer homomorphism is not exactly rocket science! Since $[G,G]$ is in the kernel of $\phi$, it follows that $[G,G]$ has exponent at most $n$, and since $[G,G]$ is finitely generated, it follows that it is finite. – Derek Holt Jun 21 '14 at 11:54
• @DerekHolt: Almost everything is elemantary for you, Derek. – mesel Jun 21 '14 at 11:56
• @DerekHolt: What I mean we have $x^ny^n=(xy)^n$, and Is it it really necassary to define new concept or function to prove this ? Can't we show this by using properties of $Z(G)$, or usual elementwise operation in a group ? Of course, the answer maybe no. – mesel Jun 21 '14 at 12:03
• I would certainly be interested if you could, but my guess is that setting up the transfer homomorphism is the most natural and direct route to proving it. The problem is that if you try and do anything directly with $(xy)^n$ then you just a mess of commutators. – Derek Holt Jun 21 '14 at 15:16
• @NickyHekster: This sencetence is actually showing my respect to him, Nicky. I was serious about that. – mesel Jun 21 '14 at 20:24

## 1 Answer

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.

• I don't get the "$(\star),(\star)$" step; care to add a few more details? -- Also, I suspect you can replace $Z(G)$ by any subgroup of $Z(G)$? – darij grinberg Nov 29 '14 at 1:05
• Since $e(g,r)^{-1} g r$ is central, the "big" product expands to two products, one being $g^n$ because of $(\star)$ and a variable substitution $r \leftrightarrow e(h,r)$, and the other being $h^n$ because of $(\star)$. – Martin Brandenburg Nov 29 '14 at 11:45
• Your guess is right, subgroups of $Z(G)$ won't give anything more general. If $H$ is a subgroup of $Z(G)$ of index $m$, then $n$ divides $m$ and hence the map $g \mapsto g^m$ is the map $g \mapsto g^n$ composed $m/n$-times with itsself. – Martin Brandenburg Nov 29 '14 at 11:47
• Ouch! There are a few steps omitted here. Thanks for clarifying, but it wouldn't hurt to do so in the post itself. – darij grinberg Nov 29 '14 at 18:12
• Sorry, I was so immersed into the problem when I wrote up the solution that I thought that this step is trivial. I hope that you like my edit. – Martin Brandenburg Nov 29 '14 at 18:14