# $H\leq Z(G)$ for a normal Subgroup $H\leq G$ Dummit Foote 4.4.12,13,14

Title is More General (which may not be true)

But the Question is :

Let $G$ be a Group of order $3825$. Prove that if $H\unlhd G$with $|H|=17$ then $H\leq Z(G)$.

what i have done so far is $3825=17\times 3^2\times 5^2$.

As $H\unlhd G$ we see that $N_G(H)=G$ and by a result stating

$N_G(H)/C_G(H)\text {is isomorphic to a subgroup of }Aut(H)$

we see that $G/C_G(H)$ is isomorphic to subgroup of $Aut(H)$.

As $H$ is of prime order it is cyclic, So, $|Aut(H)|=16$

As $C_G(H)$ is a subgroup of $G$, $|C_G(H)|$ divides $|G|$ and with the condition $|G/C_G(H)|$ divides $16$,

i.e., some factor of $16$ divides $|G/C_G(H)|$ and $(16,3825)=1$ Thus $C_G(H)=G$ and so i conclude $H\leq Z(G)$

Please let me know if my approach can be reduced to a simpler approach.

and I would like to look for a generalization of this :

If $G$ has a Normal Group of prime order $p$ and no factor of $|G|$ divides $p-1$ then $H\leq Z(G)$.

I would like to see if this is correct? I feel it is correct But Just for a clarification.

• One other fun generalization: let $G$ be a finite group of odd order and let $N \unlhd G$ with $|N|=p$, with $p$ a Fermat prime number. Then $N \subseteq Z(G)$. – Nicky Hekster Sep 3 '13 at 10:32
• I mean, the general fact is that if $N\unlhd G$, and $\left([G:N],|\text{Aut}(N)|\right)=1$ then $N\leqslant Z(G)$. – Alex Youcis Oct 6 '13 at 19:40
Proof: Let $H=\langle a | a^p=1\rangle$, $x\in G, |x|=m, x^{-1}ax=a^r$ for some $r$. Then $x^{-p+1}ax^{p-1}=a^{r^{p-1}}=a$, since $r^{p-1}\equiv 1 (\mod p)$. Further, gcd$(m,p-1)=1$, i.e. $\alpha m +\beta (p-1)=1$ for some $\alpha,\beta$. Then $[x,a]=[x^{\beta (p-1)},a]=1$. So $a\in Z(G)$.