- #1
Artusartos
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Homework Statement
How can we prove that a subgroup H of [itex]Gl_2(Z_3)[/itex] is normal?
These are the elements of H:
[tex]\begin{pmatrix}1&1\\1&2 \end{pmatrix}[/tex]
[tex]\begin{pmatrix}1&2\\2&2 \end{pmatrix}[/tex]
[tex]\begin{pmatrix}2&1\\1&1 \end{pmatrix}[/tex]
[tex]\begin{pmatrix}2&2\\2&1 \end{pmatrix}[/tex]
[tex]\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}[/tex]
[tex]\begin{pmatrix} -1 & 0 \\0 & -1 \end{pmatrix}[/tex]
[tex]\begin{pmatrix} 0 & 2 \\1& 0 \end{pmatrix}[/tex]
[tex]\begin{pmatrix} 0 & 1 \\2 & 0 \end{pmatrix}[/tex]
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 [itex]Gl_2(Z_3)[/itex]. 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 [itex]f: Gl_2(Z_3) \rightarrow Gl_2(Z_3)/H[/itex] is a homomorphism...but I'm stuck...
We know that [itex]f(a)f(b)=aHbH[/itex] and that [itex]f(ab)=abH[/itex]. But how can we show that aHbH=abH? Can anybody please give me a hint?
Thanks in advance