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The question says,

Let $ g_1, g_2, g_3 ... g_n$ be representatives of all the distinct conjugacy classes of a finite group $G$, such that these elements pairwise commute. Prove that $G$ is abelian.

I just want my proof to get verified, as this is really simple, I'm a little sceptical about it.

Proof:

Let $C_G(g_i)$ be the centraliser of the element $g_i$. Since it is given all the $g_i$'s pairwise commute, we have $$ C_G(g_1) \cap C_G(g_2) \cap ... \cap C_G(g_n) \supset \{g_1, g_2, g_3 ... g_n\} $$.

Let us assume $|G|=N$. Since we have $|C_G(g_i)| \geq n, \forall i \in \{1,2,...,n\}$, from the class equation we have

$$ |G| = \sum_{i=1}^n{|G : C_G(g_i)|} $$ or, $$ N \geq \left( \frac{N}{n}\right) .n$$

and the equality holds iff $|C_G(g_i)|=n \forall i \in \{1,2,...,n\}$ .

Thus we have, $C_G(g_i)=\{g_1, g_2, g_3 ... g_n\} \forall i \in \{1,2,...,n\}$.

However, from the class equation, we have $C_G(g_i)=G$ for at least one $g_i$, which belongs to the centre of the group $Z(G)$.

Hence $G= \{g_1, g_2, g_3 ... g_n\}$ and the group is abelian (Proved).

Is it all right or am I missing something?

L--
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    For that matter, there is another proof: consider $H=\langle g_1, \cdots , g_n \rangle$. then $H$ is an abelian subgroup of $G$ and $\bigcup_{g \in G}H^g = G$. This implies (see for example https://math.stackexchange.com/questions/1508811/showing-that-the-conjugates-of-a-proper-subgroup-do-not-cover-the-group) that $G=H$, whence abelian. – Nicky Hekster Jun 17 '20 at 15:56
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    There is no “the representative of $Z(G)$”. In fact, each element of $Z(G)$ is in a conjugacy class by itself. The element you know for sure has cnetralizer equal to $G$ is $e$. Also, you should state somewhere in the problem that the given set is a complete set of conjugacy class representatives (that is, you have one element from each conjugacy class of $G$). – Arturo Magidin Jun 17 '20 at 16:24
  • What do you mean with the statement $C_G(g_i)\ge n$? A group cannot be larger than an integer? Or? – Jyrki Lahtonen Jun 17 '20 at 16:30
  • My bad, I meant to say the cardinality of the centralizer. – L-- Jun 17 '20 at 16:36
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    Is the statement still true if $G$ and the set of representatives of conjugacy classes in $G$ are allowed to be infinite? If $G$ is a $2$-Engel group, then it is easy to see that the statement is true, since all centralizers in $G$ are normal. Hence, any counterexample must not be a $2$-Engel group. – Geoffrey Trang Jun 18 '20 at 16:22

2 Answers2

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Consider $$N=\sum_{i=1}^n [G:C_G(g_i)]$$ The point here is that since $[G:C_G(g_i)]$ is at most $\frac Nn$ and there are $n$ terms, each term must be exactly $\frac Nn$, otherwise the two sides of the equation could not be equal. But the identity element (which must be one of the $g_i$) is centralized by the whole group, so $\frac Nn=1$, hence each conjugacy class consists of only one element and the group is abelian.

Matt Samuel
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Let $H = \langle g_1, g_2, \ldots g_r \rangle$. It is clear $H \leq C_{G}(g_i)$ for any of the representatives. Note that if $H$ is a proper subgroup, then $G \neq \bigcup_{g \,\in\, G}\, gHg^{-1}.$ (You can find how to prove this with guidance in section 4.3, question 24 of Dummit & Foote.)

Yet, for any $g \in G,\, g$ shares a conjugate class with an element of $H$. Thus, $H = G = C_{G}(g_i).$

311411
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Husam Ismael
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