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Thread: quotient group of GL2(Z/5Z) subgroups.

  1. #1
    Junior Member platinumpimp68plus1's Avatar
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    quotient group of GL2(Z/5Z) subgroups.

    G=2x2 matrixes with a1a1=m, a1a2=b, a2a1=0, a2a2=1. (where integers are viewed as residue classes in Z/5Z)

    H=2x2 matrixes with a1a1=1, a1a2=c, a2a1=0, a2a2=1.

    Prove that the quotient/factor group G/H is cyclic and of order 4.

    I'm confused because I can't see how its of order 4. I think its should be 5. Any help is appreciated!!
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  2. #2
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    Quote Originally Posted by platinumpimp68plus1 View Post
    G=2x2 matrixes with a1a1=m, a1a2=b, a2a1=0, a2a2=1. (where integers are viewed as residue classes in Z/5Z)

    H=2x2 matrixes with a1a1=1, a1a2=c, a2a1=0, a2a2=1.

    Prove that the quotient/factor group G/H is cyclic and of order 4.

    I'm confused because I can't see how its of order 4. I think its should be 5. Any help is appreciated!!
    As the title says, I'll assume G and H are subgroups of $\displaystyle GL_2(F)$, where $\displaystyle F=\mathbb{Z}/5\mathbb{Z}$ (m is not zero in this case). Here is the sketch of the proof.

    Let $\displaystyle {(\mathbb{Z}/5\mathbb{Z})}^{\times}$ be a multiplicative group of $\displaystyle \mathbb{Z}/5\mathbb{Z}$ excluding 0. Let $\displaystyle \phi:G \longrightarrow {(\mathbb{Z}/5\mathbb{Z})}^{\times}$ be a surjective group homomorphism such that $\displaystyle \phi(A) = det (A)$, where $\displaystyle A \in G$ (verify this is a well-defined group homomorphism). The kernel of $\displaystyle \phi$ is a group $\displaystyle H$. By the first isomorphism theorem, $\displaystyle |G:H | = 4 $ (The order of a multiplicative group $\displaystyle {(\mathbb{Z}/5\mathbb{Z})}^{\times}$ is 4).
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