I have a question about the following problem [Finite Group Theory, Martin Isaacs, Chapter 5]:
Let $G$ be a finite group and let $H \subset Z(G)$ such that whose index in $G$ is $n$. Then using Transfer Evaluation Lemma we know that the map from $G$ to $H$ defined by $g \rightarrow g^n$ is a homomorphism. Well, I wonder how we can show that if $|H|$ is coprime to $n$, then $G= H $ $\times$ $\ker(v)$ (i.e., G is direct product of $H$ and $ker(v)$), where $v$ is transfer map $G$ to $H$. Actually it is obvious that the intersection of these two set is trivial since $|H|$ is coprime to $n$. How can I finish this problem ?
If $|H|$ is coprime to $n=[G:H]$, and the power map $v(g)= g^n$ is a homomorphism $G\to H$, then its restriction to $H$ is an automorphism because it has an inverse function ($g\mapsto g^{n^{-1}\bmod |H|}$). Therefore, the homomorphism $v$ must be onto. Suppose the kernel is $K$. You know $K\cap H$ is trivial because $v$ is bijective restricted to $H$. To show $G=KH$, suppose $g\in G$. If it were of the form $g=hk$, then we could apply $v$ to $g$ and get $v(h)$. So simply define $h=v|_H^{-1}(v(g))$ and $k=h^{-1}g$ to conclude. As $H$ is central, it is of course normal, so the conditions for $G=K\times H$ to be an internal direct product are met.