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The set $G=\left\{\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}\mid n\in \Bbb Z\right\}$ with the operation of matrix multiplication is a group. Show that $$\phi:\Bbb Z \to G,$$ $$\phi(n)=\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}$$ is a group isomorphism (where the operation on $\Bbb Z$ is ordinary addtion).

TO show it's isomorphism: I know I must show one-to-one, onto and homomorphism. I've done these examples before but never with matrices.

How can I show if $\phi(a)=\phi(b)$ then $a=b$? Same question for onto and operation preserving with matrices.

Thank you!

user26857
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maidel b
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5 Answers5

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Hint: For your first question, write down $\phi(a)$ and $\phi(b)$ (go ahead, write down the matrices on a sheet of paper). Now, if those two are equal, what does it tell you?

The other parts are obviously different, but the idea is the same: just look at the matrices involved and use what you know about matrix multiplication (for the last part).

rogerl
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  • Thank you. I understand one-to-one after writing out the steps. I'm struggling with onto. Let me make sure I'm thinking correctly. I'm trying to find $\phi(?)=n$. Is there a different way? I'm not able to see what I'm supposed to do. – maidel b Oct 23 '15 at 17:50
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    @ShayAbbott No. You want to take an arbitrary $g\in G$ and find an $n\in Z$ so that $\phi(n)=g$. But if $g\in G$, then you know what $g$ looks like... – Teepeemm Oct 23 '15 at 18:09
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Hint:

$\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}\begin{pmatrix}1 & m \\ 0 & 1 \\ \end{pmatrix} = \begin{pmatrix}1 & n+m \\ 0 & 1 \\ \end{pmatrix}$

That should help with proving $\phi$ is a homomorphism.

Once you have proven that a homomorphism exists, you must prove it is bijective to prove the mapping is a isomorphism. You already have that the mapping in injective, you must prove it is surjective.

kleineg
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    This is what I used. Thank you! – maidel b Oct 23 '15 at 16:44
  • Glad it helped. – kleineg Oct 23 '15 at 17:31
  • If $\phi$ is a homomorphism from $\Bbb Z \to G$, shouldn't $\phi^{-1}$ take matrices as input? – pjs36 Mar 16 '17 at 19:43
  • OK, yes, that's better now – pjs36 Mar 16 '17 at 20:09
  • This is not the inverse. As in function inverse. This is not even a different function from $\phi$. What the hell? In what world the inverse of a mapping $X\to Y$ is a mapping $X\to Y$? How changing $n$ to $-n$ proves that something is bijective? Come on. The proper inverse is defined by $\begin{bmatrix}1 & n \ 0 & 1\end{bmatrix}\mapsto n$. I don't believe there are 2 answers with the same mistake. – freakish Mar 16 '17 at 20:11
  • @kleineg No. The property $f(g^{-1})=f(g)^{-1}$ is preserved by all homomorphisms. In your notation $\phi(-n)=\phi(n)^{-1}$. This doesn't imply that $\phi$ is bijective. It is bijective because it has the inverse which I've defined in my previous comment. There is no other inverse. Inverses are unique. And the inverse is a function $\phi^{-1}:G\to\mathbb{Z}$. I think you confuse $\phi^{-1}(n)$ with $\phi(n)^{-1}$. – freakish Mar 17 '17 at 16:46
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So let me straight things up, since there are multiple answer which get this incorrectly. First of all you need to show that $\phi$ is a homomorphism (for that see other answers, they are fine). Then you can show that it is an isomorphism by constructing the inverse. And by the inverse we mean the function inverse, i.e. a function

$$g:G\to\mathbb{Z}$$

such that $\phi\circ g=\mbox{id}_{G}$ and $g\circ \phi=\mbox{id}_{\mathbb{Z}}$. By general property if such inverse exists then it is a homomorphism as well and so $\phi$ is an isomorphism.

So you can easily check that

$$g\bigg(\begin{bmatrix} 1 & n\\ 0 & 1 \end{bmatrix}\bigg)=n $$

is the inverse of $\phi$.

freakish
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Just note that $$ \begin{pmatrix}1 & 1 \\ 0 & 1 \\ \end{pmatrix}^n = \begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix} $$ This implies that $\phi$ is a surjective homomorphism because exponents add when you multiply powers. It is clear that $\phi$ is injective because $\phi(n)_{12}=n$.

lhf
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Hint:

There are other ways to show that this map is one-to-one.

Use that the $ker(\phi)=0$.

Does there exist an $n$ such that the rank of $\phi(n)=\begin{pmatrix}1 & n \\ 0 & 1 \\ \end{pmatrix}$ is not $2$?

Alternatively, construct $\phi^{-1}(n)=\begin{pmatrix}1 & -n \\ 0 & 1 \\ \end{pmatrix}$. The existence of an inverse with the homomorphic property implies it's an isomorphism.

kslote1
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  • That's not the inverse. The proper inverse is a function on matrices. It maps $\begin{bmatrix}1 & n \ 0 & 1\end{bmatrix}\mapsto n$. Also if $\phi$ is a homomorphism then so is $\phi^{-1}$. So the existance of $\phi^{-1}$ is enough, no need to check that it is a homomorphism. – freakish Mar 16 '17 at 20:07