# Show that if $x\in N$, then $\left \vert x^{N} \right \vert$ divides $\left \vert x^{G} \right \vert$

Let $N$ be a normal subgroup of $G$, and $|x^{G}|$ denote the size of the conjugacy class containing $x$. Please show the following statements:

1. If $x\in N$, then $\left \vert x^{N} \right \vert$ divides $\left \vert x^{G} \right \vert$

2. If $x\in G$, then $\left \vert ( xN )^{G/N} \right \vert$ divides $\left \vert x^{G} \right \vert$

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## 1 Answer

Let $G$ act on itself by conjugation. Then $|\mathcal{O}(x)|=|x^G|$. Thus $|x^G|=[G:C_G(x)]$. So, what we need to prove is that $[N:C_N(x)]$ divides $[G:C_G(x)]$.

For the second one, let $\phi:G\rightarrow G/N$ be the canonical homomorphism. Now observe that the image of $C_G(x)$ under $\phi$ centralizes the $\phi(x)$ in $G/N$, whence $\phi[C_G(x)]\leqslant C_{G/N}(\phi(x))$.

Thus $|\phi[C_G(x)]|$ divides $|C_G(x)|$ so $|(xN)^{G/N}|=[G/N:C_{G/N}(\phi(x))]$ divides $[G/N:\phi[C_G(x)]]$ divides $[G:C_G(x)]=|x^G|.$

Now, having seen that, can you prove the first part?

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thanks , I could prove the first part ,it is a special case for the second part. – Song Dec 30 '12 at 12:11