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**davismj** Let G = $\displaystyle \bigg\{\begin{bmatrix} a & b \\ 0 & d \end{bmatrix} \colon a,b,d \in R, ad \neq 0\bigg\}$. Let N = $\displaystyle \bigg\{\begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} \colon b \in R\bigg\}.$

Then N is normal in G. Prove that G/N is abelian.

Let $\displaystyle g_1,g_2 \in G$ so $\displaystyle Ng_1Ng_2 = Ng_1g_2 \in G/N.$

I've tried a few different things, but they all come back to this N/G being not abelian. Any ideas? Thanks in advance.

For instance, we want to show that

$\displaystyle Ng_1Ng_2 = Ng_2Ng_1$

Equivalently,

$\displaystyle Ng_1g_2 = Ng_2g_1 \implies g_1g_2 = g_2g_1$

However, G is not abelian.