This is a question from Dummit & Foote.
Let G be a group of order 203. Prove that if H $\leq$ G is a normal subgroup of order 7 then H $\leq$ Z(G). Hence prove that that G is abelian.
My attempt:
H is a subgroup of prime order, so it is cyclic and hence abelian. This means $H \leq C_G(H)$. By Lagrange's theorem, $C_G(H)$ (applied to $H \leq C_G(H)$ and $ C_G(H) \leq G$), this means $C_G(H)$ is either of order 7 or 203. I am not sure how to eliminate one of the cases. If I assume it is of order 203, and hence equal to $G$, immediately means that $H \leq Z(G)$, as it commutes with all the elements of G. Using Lagrange's theorem again on $Z(G)$, you can say that $Z(G)$is either $H$ or $G$. If it is $H$, then $|G/Z(G)|$ equals 29, which implies that it is cylic, and hence G is abelian, which is a contradicts $Z(G)=H$ as $Z(G)=G$. Hence $Z(G)=G$.
So coming to the point, I want to know why $C_G(H) \ne 7$
Thanks!
Best Answer
By the Normalizer/Centralizer theorem if $H$ is a subgroup of $G$, then $$\frac{N_{G}(H)}{C_{G}(H)}\cong Im(Aut(H)).$$ Now since $H$ is normal, then $N_{G}(H)=G$. Also since $H$ is of order 7, then Aut(H) is order 6. Hence, $\mid\frac{G}{C_{G}(H)}\mid\mid 6$. But 7 is the least prime that divides the order $G$. So $\left| G\right|=\left| C_{G}(H)\right|$ and therefore $H\leq Z(G)$.