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Show that the following maps are group homomorphisms and find their kernels:

1) $\theta: \Bbb Z \rightarrow GL_2$

$\theta(n) = $$ \begin{pmatrix} 1 & n \\ 0 & 1 \\ \end{pmatrix} $

My attempt:

Let $y\in\Bbb Z$ such that

$\theta(y) = $$ \begin{pmatrix} 1 & y \\ 0 & 1 \\ \end{pmatrix} $

Then $\theta(n) \theta(y) = $$ \begin{pmatrix} 1 & y \\ 0 & 1 \\ \end{pmatrix} $$ \begin{pmatrix} 1 & n \\ 0 & 1 \\ \end{pmatrix} $$ = \begin{pmatrix} 1 & y+n \\ 0 & 1 \\ \end{pmatrix} $ = $\theta(n+y)$

So $\theta: \Bbb Z \rightarrow GL_2$ is a homomorphism. And I think ker$\theta = \theta(0)$ as \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} is the identity of the $GL_2$

(Is that an efficient enough proof?)

2) $\theta:\Bbb Q$ \ {0} $\rightarrow GL_2(\Bbb Q)$ given by $\theta(a) = \begin{pmatrix} a & 0 \\ 0 & 1 \\ \end{pmatrix} $

My attempt:

Let there exist $b \in \Bbb Q$ \ {0} such that $\theta(b) = \begin{pmatrix} b & 0 \\ 0 & 1 \\ \end{pmatrix} $

Then we have:

$\theta(a)\theta(b)= \begin{pmatrix} a & 0 \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} b & 0 \\ 0 & 1 \\ \end{pmatrix} $ = $ \begin{pmatrix} ab & 0 \\ 0 & 1 \\ \end{pmatrix} $ = $\theta(ab)$

ker$\theta= \theta(1)$

etc

Is this correct way to answer this question?

This isn't homework, by the way. I'm revising for an exam I have on monday and these questions were in our practice sheets. If you have more tips for me on my first ever Abstract Algebra exam please let me know!

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