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Math Help - Prove a Quotient Group is Abelian

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by davismj View Post
    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
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