How do I finish each line of this proof? The conjugacy classes of G partition G

Question

How do I finish each line of this proof?

The conjugacy classes of G partition G because conjugation is a ...
For each x in Z(G), the conjugacy class of x is...
Let {1, z1, z2,

The conjugacy classes of G partition G because conjugation is a ... For each x in Z(G), the conjugacy class of x is... Let {1, z1, z2,..., zm} be all of the elements of the center, Z(G). The number of elements in conjugacy classes containing these elements is... Examining the conjugacy classes for elements which are not in the center, we have ... Counting the number of elements in we have ...

Explanation / Answer

Let G be a group.

Let |G| be the order of G.

Let Z(G) be the center of G.

Let x?G.

Let NG(x) be the normalizer of x in G.

Let [G:NG(x)] be the index of NG(x) in G.

Let m be the number of non-singleton conjugacy classes of G.

Then:

|G|=|Z(G)|+?j=1m[G:NG(xj)]

Proof 1

From Conjugacy Classes of Center Elements are Singletons, all elements of Z(G) form their own singleton conjugacy classes.

Abelian Group

Suppose G is abelian.

Then from Group equals Center iff Abelian we have Z(G)=G.

So there are as many conjugacy classes as there are elements in Z(G) and hence in G.

So in this case the result certainly holds.

?

Non-Abelian Group

Now suppose G is non-abelian.

Thus Z(G)?G and therefore G?Z(G)??.

From Conjugacy Classes of Center Elements are Singletons, all the non-singleton conjugacy classes of G are in G?Z(G).

From the way the theorem has been worded, there are m of them.

Let us choose one element from each of the non-singleton conjugacy classes and call them x1,x2,

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