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

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!

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In fact,when $G/H$ is cyclic then there is an element $\bar{x}=xH\in G/H$, $G/H=\langle\bar{x}\rangle$ so $G=\langle x,H\rangle$ and so $G$ is abelian. – Babak S. Jan 25 at 14:11
@Babak Nice observation! – amWhy Jan 25 at 14:17
@BabakSorouh: I don't follow your reasoning. If a group $G$ has a cyclic normal subgroup $H$ such that $G/H$ is also cyclic, it doesn't follow that $G$ is abelian. $S_3$ is the smallest counterexample. You need to first prove that $H$ is in the center, then the argument does work. – Jyrki Lahtonen Jan 25 at 17:46
@JyrkiLahtonen: Yes exactly! I followed the way of proofing just after the OP showed $H$ is in the center. Of course, otherwise, my claim is wrong. Thanks. – Babak S. Jan 25 at 17:59
@Babak: Ok. I was a bit confused about your starting point, and wanted to make sure. Sorry. – Jyrki Lahtonen Jan 25 at 18:27

Hint: $203=7\cdot 29$, so you know that there exists a Sylow 29-subgroup $P$. Normality of $H$ means that any element $x$ of $P$ acts on $H$ by conjugation, which is an automorphism of $H$. If $x$ is of order 29, then this automorphism must have an order that is a factor of 29. What do you know about the (orders of) automorphisms of $H$?

Edit: Here is a solution without using Sylow theorems. Assume that $H$ is generated by an element $z$, so $H=\{1=z^7,z,z^2,\ldots,z^6\}$. Let $x\neq1$ be any other element of $G$. It was given that $H$ is a normal subgroup, so we know that $xzx^{-1}=z^j$ for some $j\in\{1,2,3,4,5,6\}$. Then it follows that $$x^2zx^{-2}=x(xzx^{-1})x^{-1}=xz^jx^{-1}=z^{j^2}.$$ Here we can, of course, compute $j^2$ modulo $7$. Continuing in the same way (induction, if you wish) we get that $$x^rzx^{-r}=z^{j^r},$$ for all natural numbers $r$. Again we are at liberty to compute the exponent $j^r$ modulo $7$.

You probably know (Little Fermat) that if $j\neq0$, then $j^6\equiv1\pmod{7}$. The idea is to combine that bit with our information about the order of $x$. Why is that important? If $\ell=\mathrm{ord}_G(x)$, then $x^\ell=1$, so we must have $$z=1\cdot z\cdot1=x^\ell zx^{-\ell}=z^{j^\ell}.$$ This shows that we must have $j^\ell\equiv 1\pmod 7.$

Let's take stock. We know that $j^\ell$ and $j^6$ are both congruent to 1 modulo 7. But $\ell$ has to be one of $7,29,$ or $203$. Therefore $\ell$ and $6$ have no common divisors. Therefore the order of $j$ in the group $\mathbb{Z}_7^*$ must be a factor to $\gcd(6,\ell)=1$, so we can deduce that $j\equiv 1\pmod 7$. But wait a minute! We have just shown that $xzx^{-1}=z^j=z$. This means that $x$ commutes with $z$, hence with all powers of $z$. In other words $x\in C_G(H)$. Q.E.D.

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 Could we conclude that if $|G|=pq,p By the Normalizer/Centralizer theorem if$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$H\leq Z(G)$. -  How do you know$|Aut(H)|=6$, from the order of$H$? – ramanujan_dirac Jan 25 at 18:57 The order of the automorphism group of a group of order a prime$\,p\,$is$\,p-1\,$– DonAntonio Jan 25 at 20:41 Let$\,P,Q\,$be the Sylow subgroups of order$\,7\,,\,29\,$of$\,G\, $, resp. Since they both are normal ($\,P\,$by the given info,$\,Q\,$by Sylow theorems since$\,n_{29}=1\,$), we get that their product generates their own direct product, i.e.: $$PQ\cong P\times Q$$ But we also have that $$|PQ|=\frac{|P||Q|}{|P\cap Q|}=|P||Q|=203\Longrightarrow G=PQ\cong P\times Q$$ (Read here) Since clearly$\,P\,,\,Q\,$are abelian (in fact cyclic), so is their direct product and thus also$\,G\,\$

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 Nice approach, Don +1 – Babak S. Jan 25 at 18:51 Can this be done in any other way without using Sylows theorem, because this exercise is in the section before that, and I haven't studied Sylow theorems either. Thanks in advance! – ramanujan_dirac Jan 25 at 18:54 Maryam's answer is exactly what you need, then. It doesn't use Sylow theorems at all. – DonAntonio Jan 25 at 20:43