3

this is a question from a brazillian book, Paulo A. Martin's "Grupos, Corpos e Teoria de Galois" (portuguese for "Groups, fields and Galois Theory" (The actual word field translate as "Campo", but in the math context, the term used is "Corpo", wich means "body")).

It starts with the classic old question that was answered many times in this site, in this link, only changing $p$ for $3$.

The next part is to prove the following:

Let $G$ be a group of order $3^2q$, with $q$ prime, $q>3$. If $q\equiv 1\pmod3$ and $G$ is not abelian show that $|Z(G)|=1 \ \text{or} \ 3$.

Ok, here I start to work. Right, if $G$ is not abelian, then $G / Z(G)$ is not cyclic, so, $|G / Z(G)|$ is not a prime and as $ |Z(G)|$ divides $|G|$, $|Z(G)| = 1, 3 \ \text{or} \ q$. The problem here is show that we can't have $|Z(G)|=q$. But how? How can I prove this? Is this right? I got some things, like $n_{q}=1$ and $n_{3}=1 \ \text{or} \ q$ but I don't know if this is usefull.

I appreciate any help! Thanks in advance for everyone who read!

Arturo Magidin
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big_GolfUniformIndia
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2 Answers2

3

If $|Z(G)|=1$, then we are done.

If $q$ divides $|Z(G)|$, then Sylow-$q$ will be central subgroup, so $G$ is direct product of Sylow-$3$ and Sylow-$q$ (why?), so it will be abelian.

If $q$ does not divide $|Z(G)|$ but $3$ divides $|Z(G)|$, we try to show, $3^2$ does not divide $|Z(G)|$. If this is not the case, then taking $K$ to be a central subgroup of order $3^2$, we get that $G/K$ is cyclic, so $G$ is abelian.

Similarly, we can show that if $3q$ divides $|Z(G)|$ then $G$ is abelian.

This covers all the basic and sufficient cases.

Maths Rahul
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  • Great! I've only did not get that if $q$ divides order of center, then $G$ is the direct product. I can't remember any result that implies it. There is some source or some text where I can find this result? Thanks again! – big_GolfUniformIndia Apr 13 '21 at 08:23
  • Suppose $q$ divides order of center. This implies Sylow-$q$ is central (clear?). What about Sylow $3$? We know (it is abelian and) intersects trivially with Sylow $q$. But Sylow $q$ is central, so it commutes with all members of Sylow $3$. Now do you get the lines? – Maths Rahul Apr 13 '21 at 10:38
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    Don't you need that $q \equiv 1$ mod $3$? – Nicky Hekster Apr 13 '21 at 13:19
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    No, this is not necessary. Here is a more elementary proof. Suppose that $|Z(G)|$ is divisible by $q$. Let $x$ have order 3, and consider $C_G(x)$. Since the Sylow $3$-subgroup has order $9$, it is abelian, whence $9\mid |C_G(x)|$. Since $q\mid |Z(G)|$, we have $q\mid |C_G(x)|$. Thus $x$ is central. As all elements of order $3$ are central, and $G/Z(G)$ cannot be non-trivial and cyclic, we see that $G=P\times Q$ for $P$ and $Q$ Sylow $3$- and $q$-subgroups. – David A. Craven Apr 13 '21 at 19:38
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This argument doesn't use any Sylow theorem.

In principle (Lagrange), nonabelian $G$ with $|G|=3^2q$ can have center of order any among $1,3,3^2,q,3q$. Hereafter, $\mathcal{K}:=\{\text{noncentral conjugacy classes of}\space G\}$, $k:=\#\mathcal{K}$ and $\mathcal{R}$ is a complete set of representatives of the elements of $\mathcal{K}$. Note that $\mathcal{K}\ne\emptyset$ by assumption on $G$.

  • If $|Z(G)|=3^2$, then for every noncentral $g\in G$, $|C_G(g)|=3^2$, because $3^2\nmid q$; therefore, all the noncentral conjugacy classes comprise exactly $q$ elements and their (disjoint) union must count $|G|-|Z(G)|=3^2q-3^2=3^2(q-1)$ elements: contradiction, because $q\nmid 3^2(q-1)$.
  • If $|Z(G)|=q$, then, for every noncentral $g\in G$, $|C_G(g)|=3^{\alpha(g)}q$, where $\alpha(g)\in\{0,1\}$. The Class Equation, $\sum_{g\in\mathcal{R}}\frac{|G|}{|C_G(g)|}=|G|-|Z(G)|$, then yields: $$3(3l_0+l_1)=8q \tag 1$$ where $l_0+l_1=k$. By taking "$\text{mod 3}$" both sides of $(1)$, we get: $8q\equiv 0\pmod 3$, in contradiction with the assumption $q\equiv 1\pmod 3$.
  • If $|Z(G)|=3q$, then $G/Z(G)$ is cyclic, whence $G$ is abelian: contradiction.
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    The $3^2$ can be treated like the $3q$ case, as $q$ is prime. Otherwise I like this method, +1. – user1729 May 08 '21 at 14:22
  • @user1729 Thank you. Frankly, I was rather aiming to address the $3q$ case like the others, so as to give a full (Sylow- and) $G/Z(G)$-free argument. But in that way I can come up just to a $k=2q$, which doesn't seem to lead to any contradiction. If you have any ideas to "close the loop", I'd like to read it. –  May 10 '21 at 09:36
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    I'm not sure. I think the $G/Z(G)$ method is the slickest way though, as it doesn't require any messing about with specific numbers (unlike Sylow). – user1729 May 10 '21 at 09:48