0
$\begingroup$

Could someone please verify my following proof?

If group $G$ is abelian and $H\triangleleft G$, then quotient group $G/H$ is abelian.

Proof: Let $G$ be abelian and let $G/H=\left \{ gH:g\in G \right \}$. Let $g_{1}H,g_{2}H\in G/H$. Then $(g_{1}H)(g_{2}H)=g_{1}g_{2}H=g_{2}g_{1}H$ ($G$ is abelian) $=(g_{2}H)(g_{1}H)$. Therefore, $G/H$ is abelian. $\square$

$\endgroup$
3
  • 1
    $\begingroup$ Looks good to me. However I'm not aware of that notation for $H$ normal in $G$. Typically I would denote this that by $H \trianglelefteq G$. Always neat to learn about new notation. $\endgroup$ Mar 8 '18 at 4:41
  • $\begingroup$ @XandruMifsud Sorry if it confused you. I actually prefer the notation you use but I am so used to using a basic triangle from my university class! $\endgroup$
    – user482939
    Mar 8 '18 at 4:46
  • $\begingroup$ No problem, cheers! :-) $\endgroup$ Mar 8 '18 at 5:13
2
$\begingroup$

The proof is right. For a similar proof in slightly different language, you can observe that the natural group homomorphism $\varphi:G\to G/H$ is surjective, so that for every $h_1,h_2\in G/H$ we can find $g_1,g_2\in G$ so that $\varphi(g_1)=h_1,\varphi(g_2)=h_2$, from which it follows that $$h_1h_2=\varphi(g_1)\varphi(g_2)=\varphi(g_1g_2)=\varphi(g_2g_1)=\varphi(g_2)\varphi(g_1)=h_2h_1.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy