# Proof verification: If group $G$ is abelian, then quotient group $G/H$ is abelian.

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. $$\square$$

• 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. Mar 8 '18 at 4:41
• @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!
– user482939
Mar 8 '18 at 4:46
• No problem, cheers! :-) Mar 8 '18 at 5:13

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 $$h_1h_2=\varphi(g_1)\varphi(g_2)=\varphi(g_1g_2)=\varphi(g_2g_1)=\varphi(g_2)\varphi(g_1)=h_2h_1.$$