Hw10 Problem 7

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### 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

$\{x\in G |\phi(x) = \phi(a)\}$

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

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

$=\{x\in G |xa^{-1}\in H\}$

$=\{x\in G |x\in Ha\}$

$= Ha$.