Question

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

If ${\mathit{C}}_{\mathbf{i}}$ is any conjugacy class in G , prove that ${\mathit{C}}_{\mathbf{i}}\mathbf{\subseteq }\mathit{N}$ or ${\mathit{C}}_{\mathbf{i}}\mathbf{\cap }\mathit{N}\mathbf{=}\mathit{\Phi }$.

Short answer

It is proved that,$C_i\subseteq N$or $C_i\cap N=\Phi$.

Step-by-step solution

Given information

It is given that N is a normal subgroup of G , $a\in G$, C is the conjugacy class of a in G and $C_i$ is any conjugacy class in G .

Prove the statement

If $C_i\cap N=\Phi$, then there is nothing to prove.

If $C_i\cap N$is non-empty, then part (a) implies that $C_i\subseteq N$.

It has been concluded that,$C_i\subseteq N$ or $C_i\cap N=\Phi$.