Let $G$ be a finite group and let $a$ be an element of $G$. Then, $|cl ~(a)| = |G:C(a)|$

Question

Let $G$ be a finite group and let $a$ be an element of $G$. Then, $|cl ~(a)| = |G:C(a)|$ where $cl ~(a)$ refers to the conjugacy class of $a$.

The proof in the book which I am reading asks to prove the following steps :

$(i)$ Consider the function $T$ that sends the coset $g ~C(a)$ to the conjugate $gag^{-1}$ of $a$.

Hence, lets consider a function $\Psi : G/C(a) \rightarrow cl(a)$ such that

$\Psi(g ~C(a)) = gag^{-1}$

$(ii)$ Proving that $T$ is well defined

Suppose $g_1~C(a) = g_2~C(a)$ . Then we should prove that $\Psi(g_1~C(a)) = \Psi((g_2~C(a))$ for $T$ to be well defined

$g_1~C(a) = g_2~C(a) \implies g_1 g_2^{-1} \in C(a) \implies g_1 = k~g_2~~|~~k \in C(a)$

Hence, $\Psi(g_1C(a)) = g_1ag_1^{-1} = k~g_2 a~g_2^{-1}k^{-1}$

We need to show the above is $= g_2ag_2^{-1}$. We know that $a k=ka$

How do I proceed ahead here?

$(iii)$ Proving the $T$ is one-one and onto

Suppose $\Psi(g_1C(a)) = \Psi((g_2~C(a)) \implies g_1ag_1^{-1} = g_2ag_2^{-1} $

$\implies g_2^{-1}g_1 ~a = a~g_2^{-1}g_1$

$ \implies g_2^{-1}g_1 \in C(a)$

How do I prove that $g_1=g_2$ from here?

Once, it's proved that the mapping is finite, it can be proved that the mapping is onto as well, since, $G$ is finite.

Leaving aside all these steps, I have tried to explore if a homomorphism can be obtained with the mapping $\Psi : G \rightarrow cl(a)$ whose kernel is $C(a)$

Then $\Psi(g_1g_2) = g_1g_2ag_2^{-1}g_1^{-1} \neq \Psi(g_1)\Psi(g_2) = g_1ag_1^{-1}g_2ag_2^{-1}$ in general or am I mistaken?

Thank you for your help.

Answer 1

For (ii), you got the first implication backwards: $$g_1 \, C(a) = g_2 \, C(a) \implies g_2^{-1} g_1 \in C(a) \implies (g_2^{-1} g_1) \, a = a \, (g_2^{-1} g_1) $$ $$ \implies g_1 a g_1^{-1} = g_2 a g_2^{-1} $$

For (iii), your goal in the proof of one-to-one is not to prove that $g_1=g_2$. It is to prove that $g_1 \, C(a) = g_2 \, C(a)$, and you are almost there.

For (iii), although you are correct that it is unnecessary to prove onto if the groups are finite, you should know that there is nonetheless a proof of onto that does not use finiteness. Thus, one gets that for any group $G$, finite or infinite, and for any $a \in G$, the cardinality of the conjugacy class of $a$ equals the index of $C(a)$ in $G$.

For your final quote, $cl(a)$ need not be a group, it's just a subset of $G$ in general that is not closed under the operation. And $C(a)$ need not be a normal subgroup. So no, in general you cannot expect the existence of such a homomorphism.