# If $g_1, g_2, g_3 ,..., g_n$ are representatives of conjugacy classes of a group $G$ such that the elements pairwise commute, then $G$ is abelian.

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?

• 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 math.stackexchange.com/questions/1508811/…) that $G=H$, whence abelian. Jun 17, 2020 at 15:56
• 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 $G$). Jun 17, 2020 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? Jun 17, 2020 at 16:30
• My bad, I meant to say the cardinality of the centralizer.
– L--
Jun 17, 2020 at 16:36
• 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. Jun 18, 2020 at 16:22

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.
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, $$H = G = C_{G}(g_i).$$