## Every group of order 203 with a normal subgroup of order 7 is abelian

Let $G$ be a group of order 203. Prove that if $H \leq G$ is a normal subgroup of order 7 then $H \leq Z(G)$. Prove that $G$ is abelian.

Note that $203 = 7 \cdot 29$. Now $N_G(H) = G$, so that $G/C_G(H) \leq \mathsf{Aut}(H)$. Moreover, since $H$ has prime order it is cyclic, hence abelian, and so $H \leq C_G(H)$, and thus (by Lagrange) $|G/C_G(H)|$ divides 29. Now $|\mathsf{Aut}(H)| = 6$ by Proposition 17. Since 29 and 6 are relatively prime, we have $G/C_G(H) = 1$, hence $C_G(H) = G$. Thus $H \leq Z(G)$.

Now $H \leq G$ is maximal. If $Z(G) = H$, we have $G/Z(G)$ cyclic since (by Lagrange) $|G/Z(G)|$ is prime. Then $G$ is abelian, so that $Z(G) = G$; a contradiction. Thus $Z(G) = G$, and we have $G$ abelian.

• saigonbettas  On December 8, 2010 at 5:55 pm

In the second part we said :” Now H =< G is maximal. If Z(G)= H , we have G/Z(G) cyclic since |G/Z(G)| (by Lagrange) is prime. Then G is abelian, so that Z(G)=G ; a contradiction."

Then you said:" Thus ,Z(G)=G and we have abelian."

I am confused why you said it was a contradiction, then you concluded again Z(G)=G. Is there just a mistyped!!!

• nbloomf  On December 9, 2010 at 9:44 am

The first part shows that $Z(G)$ is either $H$ or $G$. By assuming that $Z(G) = H$, we get (via the $G/Z(G)$ theorem) that $Z(G) = G$. This is a contradiction of the assumption that $Z(G) = H$. Thus, $Z(G) = G$.

• neal  On July 25, 2011 at 7:46 pm

can you explain why G/C(H) has to divide 29. this seems to be the case in the other 2 problems and I can’t recall why. thanks

• nbloomf  On August 5, 2011 at 8:37 am

This is because for finite groups $G$, $|G/N| = |G|/|N|$.

• Steven  On December 5, 2011 at 8:22 pm

Sorry, but I’m confused as well about this. Why can we say $G/C_{G}(H)$ divides 29, shouldn’t we say $G/C_{G}(H) = (7 \cdot 29)/|C_{G}(H)|$? A priori, you do not know if the size of $C_{G}(H)$ is 1, 7, 29, or 203, so for all we know $|G/C_{G}(H)| = 203$ which does NOT divide 29.

I think a less hand-wavy path would have involved considering the size of the stabilizer of $H$ in $G$ and showing that the only value that works is 203.

• nbloomf  On December 5, 2011 at 8:35 pm

Since $H$ is abelian (being of prime order) it is contained in its own centralizer.

• Steven  On December 5, 2011 at 9:45 pm

Cool! Thanks. That fact totally slipped past me.

• Gobi Ree  On December 15, 2011 at 2:01 am

I was confused by the extra-explanation compared with the previous problem. And I noticed that it was because of the ask in the comments. But actually we only need to know that $|G/C_G(H)|$ divides 203.