# Quotient groups, isomorphisms, kernels

• Apr 13th 2010, 07:45 PM
kimberu
Quotient groups, isomorphisms, kernels
Let G be the group { $
\begin{bmatrix}{a}&{b}\\{0}&{c}\end{bmatrix}
$
| a, b, c are in $Z_p$ with p a prime}
Then let K = { $
\begin{bmatrix}{1}&{b}\\{0}&{1}\end{bmatrix}
$
| b in $Z_p$}

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

(A). Prove the quotient group G/K is isomorphic to $Z_p*$ x $Z_p*$
(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| $Z_p$|, |K|=| $Z_p$|, and |G/K| = 2| $Z_p$|?

Thank you so much for any help!
• Apr 13th 2010, 07:46 PM
Drexel28
Quote:

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

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

(A). Prove the quotient group G/K is isomorphic to $Z_p*$ x $Z_p*$
(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| $Z_p$|, |K|=| $Z_p$|, and |G/K| = 2| $Z_p$|?

Thank you so much for any help!

What is $\mathbb{Z}_{p{\color{red}*}}$?
• Apr 13th 2010, 07:49 PM
kimberu
Quote:

Originally Posted by Drexel28
What is $\mathbb{Z}_{p{\color{red}*}}$?

I'm actually not sure - I assumed it was the group $Z_p$ under multiplication.

Edit: Sorry, I looked in my textbook and it says " $Z_p$*" denotes the elements in $Z_p$ that have inverses under multiplication.
• Apr 13th 2010, 07:54 PM
Drexel28
Quote:

Originally Posted by kimberu
I'm actually not sure - I assumed it was the group $Z_p$ under multiplication.

That's usually denoted $\left(\mathbb{Z}/p\mathbb{Z}\right)^{\times}$

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

P.P.S. It makes sense that $K\subseteq\ker P$ since if $(a_{ij})\in K$ then $a_{11}=a_{22}=1$ and so $P((a_{ij}))=(a_{11},a_{22})=(1,1)$ and the RHS is the identity in the codomain.
• Apr 13th 2010, 08:00 PM
tonio
Quote:

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

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

(A). Prove the quotient group G/K is isomorphic to $Z_p*$ x $Z_p*$
(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| $Z_p$|, |K|=| $Z_p$|, and |G/K| = 2| $Z_p$|?

Thank you so much for any help!

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

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

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

Tonio