# Proving Conjugacy Class Theorem with Lemma and why is it necessary

The Theorem states that:

$\left | G \right |=\left | Z(G) \right |+\sum_{g'}\frac{\left | G \right |}{\left | Z(g) \right |}$

The author goes on to state a Lemma: If $S(a)$ is the conjugacy class containing $a \in G$ and $\gamma(a) =$ number of elements in the conjugacy class $S(a)$ then $\gamma(a)=\frac{\left | G \right |}{\left | Z(a) \right |}$

He further goes on to proving two claims to prove the lemma:

If $G= g_{1}Z(a)+...+g_{\gamma}Z(a), \gamma = \gamma(a)[G:Z(a)]$

Claim 1: Any conjugate $xax^{-1}$ of $a$ , $(x \in G)$ coincides with one and only one if $g_{i} a g_{i}^{-1}=a$

Claim 2 The $\gamma$ conjugates of $a$ are distinct from one another

Though I do understand the claim 2, but I can't understand the relevance or necessity of claim1. My idea is if we can prove that each element $x$ is indeed mapped and uniquely in one $g_iZ(a)$ then the job should be done. My question is what is the "behind the scenes" relevance for claim 1

Soham

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It would be helpful if you can make the explanation basic. I am learning by myself and I am not very proficient so using "orbit stabilizer theorems etc" can be daunting for me :) Thanks for the patience – Soham Sep 4 '12 at 6:39
In your formula, I think you meant the $|Z(g)|$ outside of the summation to be $|Z(G)|$ instead. Also, $\gamma(a)= \frac{|G|}{|Z(a)|}$ instead of having $Z(g)$ in the denominator, yes? And (just to be sure) $Z(x)$ for $x\in G$ is notation for the centralizer of $x$ in $G$, yes? (the notation I am familiar with for this is $C_{G}(x)$, so I wanted to make sure) – Kirk Boyer Sep 4 '12 at 6:47
Yes, apologies for the typo. – Soham Sep 4 '12 at 6:48
Yes, I also found it a bit vague, I understood it as *one and only one $g_iZ(a)$ * if... – Soham Sep 4 '12 at 6:50
Incidentally, the "Conjugacy class theorem" here is also known as the "class equation" in some texts. – Kirk Boyer Sep 4 '12 at 7:11

Let me just restate some things. You want to show the following.

If $G$ is a group and $a \in G$, then the order of the conjugacy class of $a$ is the index $(G : Z(a))$, where $Z(a)$ is the centralizer of $a$ in $G$.

Let $\{y_i\}_{i \in I}$ be a set of left coset representatives of $Z(a)$ in $G$. Here $\#I = (G : Z(a))$. There are two claims that I think you want to make.

1. If $x \in G$ then there exists an $i \in I$ such that $xax^{-1} = y_ixy_i^{-1}$.

2. If $i \neq j$ are elements of $I$ then $y_iay_i^{-1} \neq y_jay_j^{-1}$. One could also say that the $i$ in the first claim is uniquely determined by $x$.

Perhaps what you should think is that there is a map from $I$ to the conjugacy class of $a$, sending $i$ to $y_iay_i^{-1}$. You want these two sets to have the same size. The first part says that this map is surjective, and the second part says that it is injective. You need both!

The organizing principle is, as you seem to have read, the orbit-stabilizer theorem. But ignoring that, part (1) says that to compute the conjugacy class you do not have to conjugate $a$ by every element of $G$.

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The ideas used to prove that lemma would be $(1)$ that they (the conjugates of $a$) each fall into a different coset of $Z(a)=C_{G}(a)$ and $(2)$ that the conjugates of $a$ are distinct.

It turns out that any element in the same coset of $Z(a)$ gives the same result when conjugating $a$, and so the number of cosets of $Z(a)$ is exactly equal to the number of distinct conjugates of $a$ in $G$.

Now, knowing that the cosets of a subgroup $H\leq G$ partition $G$, and that the centralizer of any element is a subgroup, we get the result of the lemma: that $\gamma(a)$ is exactly $\frac{\text{the number of elements in }G}{\text{the number of elements in one coset of }Z(a)} = \frac{|G|}{|Z(a)|}$.

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I can offer you a neat proof (imo) of this important theorem.

First, we define an action of a group $\,G\,$ on itself by conjugation: $$G\times G\to G\,\,,\,x\cdot g:=g^x:=xgx^{-1}$$

We then have (I try to write down several well known notations for different things): $$[g]=\mathcal O(g):=\{g^x\;|\; x\in G\}\,\,,\,Stab(g)=G_g:=\{x\in G\;|\;g^x=g\}=Z(g)=C_G(g)$$ Finally, we not that $$\left|\,[g]\,\right|=1\Longleftrightarrow [g]=\{g\}\Longleftrightarrow \,\forall \,x\in G\,\,,\,g^x=g\Longleftrightarrow\,\forall\,x\in G\,\,,\, xg=gx\Longleftrightarrow g\in Z(G)$$

We use now the following basic theorem of group actions: $$|\mathcal O(x)|=[G:Stab(x)]=[G:G_x]\Longrightarrow\text{in our particular case}\,,\,|\mathcal O(x)|=\left[G:Z(g)\right]=\frac{|G|}{|Z(g)|}$$ and we then get that (the sum taken over disjoint orbits, of course) $$|G|=\sum |\mathcal O(x)|=\sum_{x\in Z(G)}1+\sum_{x\notin Z(G)}|\mathcal O(x)|=|Z(G)|+\sum_{x\notin Z(G)}\frac{|G|}{|Z(x)|}$$

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I am sure, its an easy thing, but I am nit just able to connect the dots. Can you help me understand how did you arrive at $[g]:=$ {$g^x|x \in G$} – Soham Sep 4 '12 at 9:53
$\,[g]\,$ is just the orbit of $\,g\,$: the set of all the elements conjugate to it, i.e.: all the elements of the form $\,xgx^{-1}=:g^x\,$...that's all. – DonAntonio Sep 4 '12 at 9:56
Well, I'm not sure what you call "period" of a group, and the above is absolutely standard names for any group, not only cyclic ones... – DonAntonio Sep 4 '12 at 9:58
I am sorry, I completely mistook it for something else. I looked up on wiki – Soham Sep 4 '12 at 10:01
I mistook O(g) as order of g, thats why the confusion. My bad – Soham Sep 4 '12 at 10:02