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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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1 Answer 1

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.

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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
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