# Prove a Quotient Group is Abelian

• Jun 3rd 2010, 11:59 AM
davismj
Prove a Quotient Group is Abelian
Let G = $\bigg\{\begin{bmatrix} a & b \\ 0 & d \end{bmatrix} \colon a,b,d \in R, ad \neq 0\bigg\}$. Let N = $\bigg\{\begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} \colon b \in R\bigg\}.$

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

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

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

$Ng_1Ng_2 = Ng_2Ng_1$

Equivalently,

$Ng_1g_2 = Ng_2g_1 \implies g_1g_2 = g_2g_1$

However, G is not abelian.
• Jun 3rd 2010, 06:04 PM
tonio
Quote:

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

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

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

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

$Ng_1Ng_2 = Ng_2Ng_1$

Equivalently,

$Ng_1g_2 = Ng_2g_1 \implies g_1g_2 = g_2g_1$

However, G is not abelian.

Hints:

1) If $G$ is any group and $N\lhd G$ , then $G/N$ is abelian iff $G':=[G:G]\leq N$

2) Calculate a general commutator $[A,B]=A^{-1}B^{-1}AB\,,\,\,A,B\in G$

3) Deduce that $G'\leq N$

Tonio