Question

Let N be a normal subgroup of G, $\mathit{a}\mathbf{\in }\mathit{G}$, and C the conjugacy class of a in G.

Use the class equation to show that $\left|N\right|=\left|C_1\right|+...+\left|C_k\right|$, where${\mathit{C}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{C}}_{\mathbf{k}}$ are all the conjugacy classes of $\mathbf{G}$ that are contained in N .

Short answer

It is proved that,$\left|N\right|=\left|C_1\right|+...+\left|C_k\right|$.

Step-by-step solution

Given that

It is given that N is a normal subgroup of G ,$a\in G$, C is the conjugacy class of $a$ in G and $C_i\text{'}s$ are conjugacy classes in G that are contained in N .

Prove the statement

It has been proved that, $C_i\subseteq N$and $C_i\text{'}s$ are conjugacy classes in G.

So .$\left|N\right|=\left|C_1\right|+...+\left|C_k\right|$

It has been concluded that,$\left|N\right|=\left|C_1\right|+...+\left|C_k\right|$.