Let $G$ be a finite group, $N$ a normal subgroup such that the orders of $N$ and $G/N$ are relatively prime. Suppose $G$ has a subgroup $H$ whose order is equal to the order of $G/N$. Show that $G$ is the semidirect product of $N$ and $H$. In addition, show that if $\varphi$ is any automorphism of $G$ then $\varphi(N)=N.$

Now because the order of $H$ and the order of $N$ are relatively prime we have that $H\cap N=\{e\}$. Also because $N$ is normal in $G$ we get that $HN$ is a subgroup of $G$. Since the order of $HN$ is equal to the order of $G$, this gives us that $G=HN$. All of this tells us that $G=N\rtimes H$.

The part I am having trouble showing is that for any automorphism of $G$, $\varphi$, we have that $\varphi(N) =N$.

Any help is greatly appreciated.

  • 1
    $\begingroup$ Take an element in N. What is the order of its image under $\phi$? If it's image is $(n,h)$ in the direct product, what can $h$ be? $\endgroup$ – Mariano Suárez-Álvarez Sep 13 '15 at 2:51
  • $\begingroup$ Well its order has to be the same, because $\varphi$ is an isomorphism. Say its order is $a$. This means $(n,h)^a=(n',h^a)=(e,e)$, where $n'\in N$. So $h^a=e$, but $a \vert |N|$ and the only way the order of $h$ can divide $|N|$ is if $h=e$ because $|H|$ and $|N|$ are relatively prime and the order of $h$ must divide $|H|$. Does that sound correct? $\endgroup$ – User112358 Sep 13 '15 at 3:08
  • $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ – Mariano Suárez-Álvarez Sep 13 '15 at 3:11
  • $\begingroup$ (Please write an answer with the details) $\endgroup$ – Mariano Suárez-Álvarez Sep 13 '15 at 3:11

I would like to credit the comments made by Mariano Suarez-Alvarez in my coming up with this answer.

Let $\varphi \in \text{Aut}(G)$.

Let $n\in N$. Then $n=(n,e)$ in $G=N\rtimes H$ Now suppose $\varphi( (n,e) )=(n',h)$. Let $a$ be the order of $(n,e)$. Now because $\varphi$ is an automorphism, and therefore an isomorphism, the order of $(n', h)$ must be $a$. Therefore $(n', h)^a=(n'',h^a)=(e,e)$, where $n''\in N$.

This tells us that $h^a=e$. Now because $h\in H$ the order of $h$ must divide $|H|$, but $a \vert |N|$. Combined with the fact that $|H|$ and $|N|$ are relatively prime this tells us that the only way $h^a=e$ is if $h=e$. Thus $\varphi((n,e))=(n',e).$

This holds for all $n\in N$ so $\varphi[N] \subset N$, and therefore $\varphi[N]=N$.


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