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!

  • 2
    $\begingroup$ 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. $\endgroup$
    – Mikasa
    Jan 25 '13 at 14:11
  • 1
    $\begingroup$ @Babak Nice observation! $\endgroup$
    – amWhy
    Jan 25 '13 at 14:17
  • $\begingroup$ @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. $\endgroup$ Jan 25 '13 at 17:46
  • $\begingroup$ @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. $\endgroup$
    – Mikasa
    Jan 25 '13 at 17:59
  • $\begingroup$ @Babak: Ok. I was a bit confused about your starting point, and wanted to make sure. Sorry. $\endgroup$ Jan 25 '13 at 18:27

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 $H\leq Z(G)$.

  • $\begingroup$ How do you know $|Aut(H)|=6$, from the order of $H$? $\endgroup$
    – user23238
    Jan 25 '13 at 18:57
  • 3
    $\begingroup$ The order of the automorphism group of a group of order a prime $\,p\,$ is $\,p-1\,$ $\endgroup$
    – DonAntonio
    Jan 25 '13 at 20:41

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.

  • 2
    $\begingroup$ Could we conclude that if $|G|=pq,p<q$ such that $p$ doen't divide $q-1$ then $G$ is necessarily cyclic? Thanks. $\endgroup$
    – Mikasa
    Jan 25 '13 at 18:06
  • $\begingroup$ @Babak: Yes (presumable $p,q$ are both primes). In that case $G$ has a single Sylow $p$-subgroup, a single Sylow $q$-subgroup, and no other proper subgroups. Therefore any of the $pq-(p+q-1)$ elements not in a proper subgroup will have to generate all of $G$, so $G$ is cyclic. $\endgroup$ Jan 25 '13 at 18:31
  • $\begingroup$ Can this be done in any other way without using Sylow theorems, because this exercise is in the section before that, and I haven't studied Sylow theorems either. Thanks in advance! $\endgroup$
    – user23238
    Jan 25 '13 at 18:55
  • $\begingroup$ @ramanujan_dirac: I added a proof that recaps all the answers in a language independent of Sylow theorems. Does that fit your needs? $\endgroup$ Jan 25 '13 at 20:25
  • $\begingroup$ What do you mean by "this automorphism must have an order ... "? How is order defined for automorphisms? Is it similar to the order of an element in a group, with the group here being $Aut(H)$ with group action as composition? $\endgroup$
    – Aritra Das
    Feb 18 '18 at 14:33

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 $\,G\,$

  • $\begingroup$ Nice approach, Don +1 $\endgroup$
    – Mikasa
    Jan 25 '13 at 18:51
  • $\begingroup$ 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! $\endgroup$
    – user23238
    Jan 25 '13 at 18:54
  • $\begingroup$ Maryam's answer is exactly what you need, then. It doesn't use Sylow theorems at all. $\endgroup$
    – DonAntonio
    Jan 25 '13 at 20:43

Here is a try at a solution but I'm having trouble showing something.

Consider the center of G $Z(G)$. We know that $Z(G)\leq G$

By Lagrange's theorem $|Z(G)|$ must divide $|G|$.

So therefore the possible order of $Z(G)$ are $1, 7, 29, 203$.

If $|Z(G)|=203$ then $Z(G)=G$ and $H\leq G=Z(G)$ and $G$ is abelian.

Now $|Z(G)|\neq29$ because if $|Z(G)|=29$ then the order of the factor group $G/Z(G)$ is $|G|/|Z(G)|$ which is equal to $203/29 = 7$.

So order of $G/Z(G)$ is prime $\implies$ $G/Z(G)$ is cyclic $\implies$ $G$ is abelian $\implies G=Z(G)$ which is a contradiction since $|G|=203$.

The same argument also shows that $|Z(G)|\neq7$.

Now here is the trouble. How can I show that $|Z(G)|≠1$?

I hope someone may be able to answer. :) Thanks.

  • $\begingroup$ The OP demonstrates correctly that $H\leq Z(G)$ and $\mid H \mid=7,$ so you know that $\mid Z(G) \mid \geq 7$ $\endgroup$
    – Mike
    Dec 14 '16 at 1:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy