# 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} $$= \begin{pmatrix} 1 & y+n \\ 0 & 1 \\ \end{pmatrix} = \theta(n+y) So \theta: \Bbb Z \rightarrow GL_2 is a homomorphism. And I think ker\theta = \theta(0) as \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} is the identity of the GL_2 (Is that an efficient enough proof?) 2) \theta:\Bbb Q \ {0} \rightarrow GL_2(\Bbb Q) given by \theta(a) = \begin{pmatrix} a & 0 \\ 0 & 1 \\ \end{pmatrix} My attempt: Let there exist b \in \Bbb Q \ {0} such that \theta(b) = \begin{pmatrix} b & 0 \\ 0 & 1 \\ \end{pmatrix} Then we have: \theta(a)\theta(b)= \begin{pmatrix} a & 0 \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} b & 0 \\ 0 & 1 \\ \end{pmatrix} = \begin{pmatrix} ab & 0 \\ 0 & 1 \\ \end{pmatrix} = \theta(ab) ker\theta= \theta(1) etc Is this correct way to answer this question? This isn't homework, by the way. I'm revising for an exam I have on monday and these questions were in our practice sheets. If you have more tips for me on my first ever Abstract Algebra exam please let me know! ## 1 Answer You’re fine, if a bit clumsy, with verifying that the maps are homomorphisms, but there are some problems with the kernels. Let’s look at the first problem. To show that \theta is a homomorphism, you really should start with two arbitrary elements of \Bbb Z and verify that \theta has the homomorphism property with respect to these two elements. Let m,n\in\Bbb Z. Then$$\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.

• THANK YOU SO MUCH! I always feel like I understand the concepts but have a lot of trouble knowing which proof-language to use to verify something. Thank you, Mr. Scott! – Siyanda Nov 17 '12 at 8:34
• @Siyanda: You’re very welcome. – Brian M. Scott Nov 17 '12 at 8:43