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Let $ G $ be a group with a partition into conjugacy classes:
$$ G = [g_1] \sqcup [g_2] \sqcup \dots \sqcup [g_r], $$
where the conjugacy class of an element $ a $ is defined as
$$ [a] = \{g a g^{-1} \mid g \in G\}. $$
Suppose that for all $ i, j \in \{1, 2, \dots, r\} $, the elements $ g_i $ and $ g_j $ commute, i.e.,
$$ g_i g_j = g_j g_i. $$
Does it necessarily follow that $ G $ is generated by these elements? That is, do we have
$$ G = \langle g_1, g_2, \dots, g_r \rangle? $$

I attempted to prove this by considering an arbitrary element $ x_1 \in [g_{i_1}] $, which can be written as $ x_1 = x_2 g_{i_1} x_2^{-1} $ for some $ x_2 \in G $, and iterating this process. However, this approach does not seem to lead to a conclusive result.

How can I prove or disprove this statement? Any direction or a counterexample would be appreciated.

Samvel Safaryan
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2 Answers2

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For counterexamples in the case $r=2$ see this MathOverflow post, which describes an infinite group $G$ in which the non-identity elements are all conjugate to each other; the other conjugacy class is, of course, just the identity. This group $G$ is not generated by a single element, because then $G$ would be infinite cyclic, but every element of an infinite cyclic group is all alone in its conjugacy class.

Lee Mosher
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If $G$ is finite, your assertion is true. In general we have the following.

Lemma If $H$ is a proper subgroup of a finite group $G$, then $\bigcup_{g \in G}H^g \subsetneqq G$.

Proof Since $H$ is proper, $|G:H| \geq 2$ and the number of conjugates of $H$ equals $|G:N_G(H)|$, hence

$$|\bigcup_{g \in G}H^g| \leq |G:N_G(H)|(|H|-1)+1 \leq |G:H|(|H|-1)+1 \leq |G|-2+1=|G|-1 \lt |G|. \square$$

With the notation in your post, put $H=\langle g_1, g_2, \ldots, g_r \rangle$. Then, since the $g_i$'s are representatives of the conjugacy classes, certainly $\bigcup_{g \in G}H^g=G$. The lemma now implies $G=H$.

If in addition the $g_i$'s commute among each other, we get that $H$ and hence $G$ is abelian.

Nicky Hekster
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