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In my textbook: Gallian's Contemporary Abstract Algebra, 9TH edition, on page 196, there is a homomorphism theorem that states the following:

Let $\phi$ be a homomorphism from G to G'.

Theorem: $\phi(a)=\phi(b)$ if and only if $a\ker\ \phi=b\ker\ \phi$.

I did my proof differently from the book and was wondering if it's possible to prove it this way:

$Proof$:

Since $\ker\phi=\{x\in G:\phi(x)=e\}$ we have:
$\phi(a)=\phi(b)$
$\phi(a)\phi(x)=\phi(b)\phi(x)$ (multiply both sides by $\phi(x)$
$\phi(ax)=\phi(bx)$ (property of homomorphism)
$a\ker\ \phi=b\ker\ \phi$. (This last part cannot be concluded, correct? Since $\phi$ is not a bijection.)

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Hint:

$$a(\ker\phi)=b(\ker\phi)\iff ab^{-1}\in\ker\phi\iff\phi(ab^{-1})=e_{G'}$$

Now, use that the homomorphism properties of $\phi$ and conclude.

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    $\begingroup$ Hi, thank you for the proof, but that is how the textbook proves it. My question, written in bold, asks if the last part of my derivation is incorrect. $\endgroup$ – user462561 Mar 20 '18 at 21:16
  • $\begingroup$ @LiftusDominatus: How do you go from $\phi(ax)=\phi(bx)$ to $a(\ker\phi)=b(\ker\phi)$? Do note that the latter doesn't necessarily mean that $ax=bx$ for all $x\in\ker\phi$ but rather means that given $a,b\in G$, for each $x\in\ker\phi$, there exists $y\in\ker\phi$ such that $ax=by$. $\endgroup$ – Prasun Biswas Mar 20 '18 at 21:30

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