Hw10 Problem 7

### Problem 7

Let $\phi : G\rightarrow G'$ be a homomorphism with kernel $H$ and let $a\in G$.
Prove the set equality $\{x\in G |\phi(x) = \phi(a)\} = Ha$

Solution

Let $e'$ be the identity of $G'$. Then

$\{x\in G |\phi(x) = \phi(a)\}$
$= \{x\in G |\phi(x)\phi(a)^{-1} = e'\}$
$=\{x\in G |\phi(xa^{-1})= e'\}$
$=\{x\in G |xa^{-1}\in H\}$
$=\{x\in G |x\in Ha\}$
$= Ha$.