# Question about $G/Z(G)$

If $G/Z(G)$ are the cosets of $Z(G)$ and the disjoint union of conjugacy classes such that $|G| = |Z(G)| + n_1 + n_2 + n_3 +\cdots+n_r$ where $n_1,n_2,\ldots,n_r$ are the number of element in the conjugacy classes. My question is since $G/Z(G)$ are cosets, shouldn't the size of each conjugacy classes and $Z(G)$ be equal? What is my misunderstanding? Maybe I am just misunderstanding the notation $G/Z(G)$ to be a quotient group?

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The size of the conjugacy class of $x$ is equal to the index of the centralizer of $x$; the centralizer of $x$ is $$C_G(x) = \{g\in G\mid gx=xg\}.$$ Now, certainly, $Z(G)\subseteq C_G(x)$ for all $x$, but there may be elements for which the centralizer is larger. For example, $C_G(e) = G$, so unless $G$ is abelian, the centralizer of the identity is strictly larger than the center.

So you can say that the size of the conjugacy class of $x$ is bounded above by the index of $Z(G)$, but it need not be equal to it.

The conjugacy class of $x$ has one element if and only if $C_G(x)=G$, if and only if $gx=xg$ for all $g\in G$, if and only if $x\in Z(G)$. So the elements of $Z(G)$ are precisely the elements whose conjugacy class is equal to just themselves. Every other element has more than $1$ conjugate, and at most $[G:Z(G)]$ conjugates.

So if you partition $G$ into conjugacy classes,you will have $|Z(G)|$ classes with exactly one element, and then a bunch of other classes with more than one element. The class equation "bundles up" all the classes with one element together into a summand that is $|Z(G)|$; the rest of the sizes of the other classes are then added.

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if I want to show $|G|/|Z(G)|$ is never a prime, which theorem is most helpful? Is it possible to show this using class equation? Originally I tried using cosets, but from your explanation I don't think it's related to cosets. – Mark Feb 13 '12 at 19:59
@Mark Since it is prime, it is cyclic. Now, by merely writing out what it means, you must be through. – user21436 Feb 13 '12 at 20:03
@Mark: Prove the following theorem: If $N\leq Z(G)$, and $G/N$ is cyclic, then $G$ is abelian. – Arturo Magidin Feb 13 '12 at 20:06
Or the perhaps more simple statement that if $x$ is non-central then $Z(G) < C_G(x) < G$ so $Z(G)$ is never maximal. – Tobias Kildetoft Feb 13 '12 at 20:21
Isn't it true that size of each conjugacy class is bounded above by the index of center, because $|Z(G)| \le |C_G(g)|$? – user21436 Feb 13 '12 at 20:22

I don't quite get the connection right:

Here is my go in putting things:

• Since $G/Z(G)$ is the group of cosets (translates) of $Z(G)$ in $G$. So, it is perfectly true that the size of each coset of $Z(G)$ in $G$ is the same.

• Now, this has nothing to do with the conjugation action of $G$ on itself. Note that it is not true that the size of each conjugacy class is the same as the size of $Z(G)$.

For instance, An Abelian group, where centre $Z(G)$ is the whole group, note that, each conjugacy class is a singleton.

More generally, the conjugacy class of each element in $Z(G)$ for any group is unity. In particular, there is always one conjugacy class of size $1$ while it is not true that $|Z(G)|=1$ in these cases.

Illustration:

Let $a \in Z(G)$. Define $\phi_a:G \to G$ by $\phi_a(g)=aga^{-1}$. That is, $\phi_a$ is the conjugation by $a$. Note that, as $a \in Z(G)$, $\phi_a(g)=aa^{-1}g=g$. That goes to say, $\phi_a$ is identity on the whole of $G$.

Now further, we can talk about the conjugacy action here.

Let $G$ act on itself by Conjugation. So, I mean the following: \begin{align*}\operatorname{Act}:G \times G &\to G\\(a,g) &\mapsto aga^{-1}\end{align*} The orbits of this action is the conjugacy classes.

Now, what is the size orbit,$|\mathcal O_g|$ of an element, $g$ in $G$?

You must have known (or prove that!) $$|\mathcal O_g|=\dfrac{|G|}{|\operatorname{stab}(g) |}$$

What is the stabilizer of a point? \begin{align*}\operatorname{Stab} (g)&=\left\{x \in G|xgx^{-1}=g\right\}\\&=\text{Set of elements in }G\text{ that commute with }g \in G\text{ }\\ &=C_G(g)\end{align*}

Now, $Z(G)= \bigcap_{g \in G} C_G(g)$. So, the $|Z(G)| \le |C_G(g)| \forall g \in G$

It is in fact true that, the size of each conjugacy class is atmost the index of $Z(G)$.

Further, that the conjugacy classes need not be of the same size is what led to the very first attempt of the classification of finite groups in the form of Class Equation !

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