Proving that a subgroup is normal.

[Artusartos]

Homework Statement

How can we prove that a subgroup H of Gl_2(Z_3) is normal?

These are the elements of H:

\begin{pmatrix}1&1\\1&2 \end{pmatrix}

\begin{pmatrix}1&2\\2&2 \end{pmatrix}

\begin{pmatrix}2&1\\1&1 \end{pmatrix}

\begin{pmatrix}2&2\\2&1 \end{pmatrix}

\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}

\begin{pmatrix} -1 & 0 \\0 & -1 \end{pmatrix}

\begin{pmatrix} 0 & 2 \\1& 0 \end{pmatrix}

\begin{pmatrix} 0 & 1 \\2 & 0 \end{pmatrix}

So the determinant of the elements is 1 and the trace is zero...and H additionally contains the identity element and the -identity element.

Homework Equations

The Attempt at a Solution

I don't think I can find the left and right cosets and show that they are equal, because there are too many elements in Gl_2(Z_3). There is a theorem in our textbook that states,

Let H be a subgroup of G. Then H is normal in G if and only if there is a group structure on the set G/H of left cosets of H with the property that the canonical map π : G → G/H is a homomorphism. If H is normal in G, then the group structure on G/H which makes π a homomorphism is unique: we must have gH · gH = ggH for all g, g ∈ G. Moreover, the kernel of π : G → G/H is H. Thus, a subgroup of G is normal if and only if it is the kernel of a homomorphism out of G.

So I'm trying to use this and show that f: Gl_2(Z_3) \rightarrow Gl_2(Z_3)/H is a homomorphism...but I'm stuck...

We know that f(a)f(b)=aHbH and that f(ab)=abH. But how can we show that aHbH=abH? Can anybody please give me a hint?

Thanks in advance

 

[jbunniii]

All the elements of $H$ have determinant $1$, so H is a subgroup of $SL_2(Z_3)$. How many subgroups of order 8 does $SL_2(Z_3)$ have?

 

[Artusartos]

All the elements of $H$ have determinant $1$, so H is a subgroup of $SL_2(Z_3)$. How many subgroups of order 8 does $SL_2(Z_3)$ have?

Thanks, but I didn't read about SL_2(Z_3) yet.

 

[jbunniii]

$SL_2(Z_3)$ is the subgroup of $GL_2(Z_3)$ consisting of the matrices with determinant 1. What I have in mind is to try the following argument: (1) $H$ is characteristic in $SL_2(Z_3)$; (2) $SL_2(Z_3)$ is normal in $GL_2(Z_3)$; (3) therefore...?

In order to make that argument, you need to know the orders of $GL_2(Z_3)$ and $SL_2(Z_3)$. Can you calculate these?

 

[Artusartos]

$SL_2(Z_3)$ is the subgroup of $GL_2(Z_3)$ consisting of the matrices with determinant 1. What I have in mind is to try the following argument: (1) $H$ is characteristic in $SL_2(Z_3)$; (2) $SL_2(Z_3)$ is normal in $GL_2(Z_3)$; (3) therefore...?

In order to make that argument, you need to know the orders of $GL_2(Z_3)$ and $SL_2(Z_3)$. Can you calculate these?

Alright, thanks