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Thread: Quotient groups, isomorphisms, kernels

  1. #1
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    Quotient groups, isomorphisms, kernels

    Let G be the group {$\displaystyle
    \begin{bmatrix}{a}&{b}\\{0}&{c}\end{bmatrix}
    $ | a, b, c are in $\displaystyle Z_p$ with p a prime}
    Then let K = {$\displaystyle
    \begin{bmatrix}{1}&{b}\\{0}&{1}\end{bmatrix}
    $ | b in $\displaystyle Z_p$}

    The map P: G --> $\displaystyle Z_p*$ x $\displaystyle Z_p*$ is defined by
    P( $\displaystyle
    \begin{bmatrix}{a}&{b}\\{0}&{c}\end{bmatrix}
    $ ) = (a, c)


    (A). Prove the quotient group G/K is isomorphic to $\displaystyle Z_p*$ x $\displaystyle Z_p*$
    (B). Find the orders |g|, |K|, and |G/K|

    my thoughts -
    (A) I've already shown that K is normal in G, and I know this would be true if K were the kernel of P, but I don't know how to show that.
    (B), I'm not sure -- is it |G|= 3|$\displaystyle Z_p$|, |K|=|$\displaystyle Z_p$|, and |G/K| = 2|$\displaystyle Z_p$|?

    Thank you so much for any help!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kimberu View Post
    Let G be the group {$\displaystyle
    \begin{bmatrix}{a}&{b}\\{0}&{c}\end{bmatrix}
    $ | a, b, c are in $\displaystyle Z_p$ with p a prime}
    Then let K = {$\displaystyle
    \begin{bmatrix}{1}&{b}\\{0}&{1}\end{bmatrix}
    $ | b in $\displaystyle Z_p$}

    The map P: G --> $\displaystyle Z_p*$ x $\displaystyle Z_p*$ is defined by
    P( $\displaystyle
    \begin{bmatrix}{a}&{b}\\{0}&{c}\end{bmatrix}
    $ ) = (a, c)


    (A). Prove the quotient group G/K is isomorphic to $\displaystyle Z_p*$ x $\displaystyle Z_p*$
    (B). Find the orders |g|, |K|, and |G/K|

    my thoughts -
    (A) I've already shown that K is normal in G, and I know this would be true if K were the kernel of P, but I don't know how to show that.
    (B), I'm not sure -- is it |G|= 3|$\displaystyle Z_p$|, |K|=|$\displaystyle Z_p$|, and |G/K| = 2|$\displaystyle Z_p$|?

    Thank you so much for any help!
    What is $\displaystyle \mathbb{Z}_{p{\color{red}*}}$?
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    What is $\displaystyle \mathbb{Z}_{p{\color{red}*}}$?
    I'm actually not sure - I assumed it was the group $\displaystyle Z_p$ under multiplication.

    Edit: Sorry, I looked in my textbook and it says "$\displaystyle Z_p$*" denotes the elements in $\displaystyle Z_p$ that have inverses under multiplication.
    Last edited by kimberu; Apr 13th 2010 at 06:51 PM. Reason: updated definition
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kimberu View Post
    I'm actually not sure - I assumed it was the group $\displaystyle Z_p$ under multiplication.
    That's usually denoted $\displaystyle \left(\mathbb{Z}/p\mathbb{Z}\right)^{\times}$

    Anyways, if you prove that this $\displaystyle P$ is an epimorphism it follows that $\displaystyle K=\ker P$ and so $\displaystyle G/K\cong \left(\mathbb{Z}/p\mathbb{Z}\right)^{\times}\times\left(\mathbb{Z}/p\mathbb{Z}\right)^\times$

    P.P.S. It makes sense that $\displaystyle K\subseteq\ker P$ since if $\displaystyle (a_{ij})\in K$ then $\displaystyle a_{11}=a_{22}=1$ and so $\displaystyle P((a_{ij}))=(a_{11},a_{22})=(1,1)$ and the RHS is the identity in the codomain.
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  5. #5
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    Red face

    Quote Originally Posted by kimberu View Post
    Let G be the group {$\displaystyle
    \begin{bmatrix}{a}&{b}\\{0}&{c}\end{bmatrix}
    $ | a, b, c are in $\displaystyle Z_p$ with p a prime}
    Then let K = {$\displaystyle
    \begin{bmatrix}{1}&{b}\\{0}&{1}\end{bmatrix}
    $ | b in $\displaystyle Z_p$}

    The map P: G --> $\displaystyle Z_p*$ x $\displaystyle Z_p*$ is defined by
    P( $\displaystyle
    \begin{bmatrix}{a}&{b}\\{0}&{c}\end{bmatrix}
    $ ) = (a, c)


    (A). Prove the quotient group G/K is isomorphic to $\displaystyle Z_p*$ x $\displaystyle Z_p*$
    (B). Find the orders |g|, |K|, and |G/K|

    my thoughts -
    (A) I've already shown that K is normal in G, and I know this would be true if K were the kernel of P, but I don't know how to show that.
    (B), I'm not sure -- is it |G|= 3|$\displaystyle Z_p$|, |K|=|$\displaystyle Z_p$|, and |G/K| = 2|$\displaystyle Z_p$|?

    Thank you so much for any help!

    For the above to be true it must be that $\displaystyle G:=\left\{\begin{pmatrix}a&b\\0&c\end{pmatrix}\;\; ;\;a,b,c\in\mathbb{Z}_p\,,\,\,ac\neq 0\right\}$ ; Now it's clear the map P is a group epimorphism and its kernel is K since $\displaystyle (1,1)$ is the unit element in the group $\displaystyle \mathbb{Z}^{*}_p\times \mathbb{}Z^{*}_p$ , and for the orders:

    $\displaystyle |G|=(p-1)p(p-1)=p(p-1)^2\,,\,\, |K|=p\,,\,\, \left|G\slash K\right|=(p-1)^2$.

    Finally: yes, $\displaystyle \mathbb{Z}^{*}_p=\left(\mathbb{Z}\slash p\mathbb{Z}\right)^{*}=\mathbb{Z}_p-\{0\}=$ the multiplicative group of the field $\displaystyle \mathbb{Z}_p$

    Tonio
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