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!