# Thread: Quotient groups, isomorphisms, kernels

1. ## 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!

2. Originally Posted by kimberu
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 $\displaystyle \mathbb{Z}_{p{\color{red}*}}$?

3. Originally Posted by Drexel28
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.

4. Originally Posted by kimberu
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.

5. Originally Posted by kimberu
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