Let $G$ a simple group. Suppose exists a conjugacy class $C$ in $G$ such that $C \neq\{e\}$ is finite. Prove $G$ is finite.

Question

So I am pretty sure my proof lacks something because I never used the fact G is simple, and I would love some help.

So this is my proof:

Let's observe the action $G \mapsto C$ by conjugation. So, $\forall g \in G$ we have $g.c = gcg^{-1}$.

Let $c_1 \in C$, and we can notice that $|Oc_1|$ is actually the size of the conjugacy class of $c_1$, because the orbit of $c_1$ is all the objects in $C$ that $c_1$ can move to using conjugation under $G$. Thus implementing $|Oc_1|=|C|$.

From the datum $C$ is final we can deduce the orbit of $c_1$ is final.

It is given $C \neq$ {e} and therefore exists $e \neq c_2 \in C$.

By invalidation suppose $e \in C$, so there exists $g \in G$ such that $gc_2g^{-1} = e \to gc_2 = g \to c_2 = e$ and that is a contradiction.

Thus implementing $e \notin C$. Therefore we have $\forall c \in C$ we have $G_C$, the stabilizer of $C$ is also finite. Thus implementing $\forall c \in C$ we have $|G_C||O_c|$ is finite and by the orbit-stabilizer theorem we have $G$ is finite.

Answer 1

This is heavily inspired by Mark's comments. Its simple to verify that the map $\phi_g:C\to C, c\to gcg^{-1} $ its bijective i.e. $\phi_g\in S_C$, hence we can define homomorphism: $$F:G\to S_C, g\to \phi_g \;\;\;(\text{verify that this is really a homomorphism}). $$ Now $ker F\unlhd G$ and because G is simple we get that $ker F\in\{\{e\},G\}$.

1)$ker F=\{e\}$ : By the first isomorphism theorem we get: $$G\cong G/\{e\}\cong F(G)\le S_C \implies |G|\le|S_C|<\infty .$$

2)$ker F=G$ : In this case $\phi_g=Id$ for every $g\in G$ hence: $$c=\phi_g(c)=gcg^{-1}\;\; \forall c\in C, g\in G$$ therefore $|C|=1$ (because $C$ its a Conjugacy class of some element $c_0$ but $gc_0g^{-1}=c_0$ everytime so $C=\{c_0\}$) and $c_0\in Z(G)$ hence $<c_0>\unlhd G\stackrel{C\neq\{e\}}{\implies} G=<c_0>$ (because $G$ its simple). Now $<c_0>$ must be finite, if not we have that $$ \{0\}\subsetneq <c_0^2>\subsetneq <c_0>=G $$ $$\text{because if }\; c_0=c_0^{2k}\implies c_0^{2k-1}=e\implies <c_0>=G \text{ finite}.$$ But now we have a proper, non trivial, normal subgroup of G and it's in contraddiction with simple assumption.

Answer 2

The proof goes as follows.

Let $G$ be a simple group with a non-trivial finite conjugacy class $C$. In consequence, $\exists x \in C$ such that $x \neq e$. Its centralizer $C_G(x)$ is a normal subgroup of $G$. As $G$ is simple, its normal subgroups can only be the trivial group $\{e\}$ or $G$ itself.

If $C_G(x) = \{e\}$, the orbit-stabilizer theorem leads straightforwardly to the desired result; indeed, we get : $$ |C| = [G:C_G(x)] = \frac{|G|}{|C_G(x)|} = |G| \quad\Longrightarrow\quad |G| = |C| < \infty. $$

The case $C_G(x) = G$ is a bit more complicated. Indeed, in that case, the center of $G$ is not trivial, since $x \in Z(G) \lhd G$. If $Z(G) \neq G$, then $G$ is not simple anymore and the proof is finished ab absurdo. If $Z(G) = G$, then $G$ is abelian and thus not simple as before, unless $G$ is a prime-order cyclic group, hence finiteness.

Answer 3

$G$ acts transitively by conjugation on $C$. Since $G$ is simple, the action is faithful. Therefore, $G$ embeds into $S_C$, which is finite because $C$ is finite. So, $G$ is finite.