Show that each of the following are homomorphisms 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!
 A: You’re fine, if a bit clumsy, with verifying that the maps are homomorphisms, but there are some problems with the kernels.
Let’s look at the first problem. To show that $\theta$ is a homomorphism, you really should start with two arbitrary elements of $\Bbb Z$ and verify that $\theta$ has the homomorphism property with respect to these two elements.

Let $m,n\in\Bbb Z$. Then $$\theta(m+n)=\pmatrix{1&m+n\\0&1}=\pmatrix{1&m\\0&1}\pmatrix{1&n\\0&1}=\theta(m)\theta(n)\;,$$ so $\theta$ is a homomorphism.

Your statement that $\ker\theta=\theta(0)$ doesn’t make sense: $\ker\theta$ is by definition a subset of $\Bbb Z$, while $\theta(0)$ is an element of $GL_2$, so they can’t possibly be equal. Go back to the definition:
$$\ker\theta=\left\{n\in\Bbb Z:\theta(n)=\pmatrix{1&0\\0&1}\right\}\;,$$
since $\pmatrix{1&0\\0&1}$ is the identity element of $GL_2$. Now you can argue as follows.

Suppose that $n\in\ker\theta$; then $\theta(n)=\pmatrix{1&0\\0&1}$ by the definition of kernel. On the other hand, $\theta(n)=\pmatrix{1&n\\0&1}$ by the definition of $\theta$, so $n=0$. Thus, $\ker\theta=\{0\}$.

You can also conclude that $\theta$ is injective (one-to-one), since its kernel is trivial, but that’s not part of the problem.
The other problem is dealt with similarly.
