# Finding smallest possible group of matrices containing a given matrix

I am trying to understand the process for solving group theory questions.

Let $a=\begin{bmatrix} 1&1\\0&1 \end{bmatrix}$ and $b=\begin{bmatrix}i&0\\0&-i\end{bmatrix}$ - 2 x 2 matrices with complex entries.

1. Describe the smallest group of 2 x 2 complex matrices containing b.
2. Describe the smallest group of 2 x 2 complex matrices containing a.

Say for question 1 - To find the smallest group I should take $b$ and multiply it by itself $n$ times until I get the identity matrix. And then the identity matrix and the $b, b^2,..., b^n$ other matrices will be the smallest group of 2x2 matrices possible? Is that the correct way to do it?

What about question 1...multiplying $a$ by itself multiple times is not going to bring me back to the identity? So what is the procedure required for that question?

• You are right that the smallest group containing $A$ will be $\langle A\rangle$, but you need to consider the possibility that a matrix $A$ need not be torsion (i.e. need not have finite order). I see that $b$ does, but $a$ does not. – anon Oct 2 '12 at 19:46
• What does <A> mean? I am trying to understand the process for answering these questions. – dukenukem Oct 2 '12 at 19:48
• It means $$\langle A\rangle=\{A^n:n\in\Bbb Z\}.$$ It forms a group under multiplication (iff $A$ is invertible) and is not necessarily finite. – anon Oct 2 '12 at 19:49
• Ok ty, so the smallest possible group containing $a$ is infinitely large? – dukenukem Oct 2 '12 at 19:50

For $b$, you are correct.

For $a$, consider the following two facts (which you can easily prove):

1. For all integers $n,m$, we have$$\begin{pmatrix}1&n\\0&1\end{pmatrix}\begin{pmatrix}1&m\\0&1\end{pmatrix}=\begin{pmatrix}1&n+m\\0&1\end{pmatrix}.$$
2. For a given integer $n$, we have $$\begin{pmatrix}1&n\\0&1\end{pmatrix}^{-1}=\begin{pmatrix}1&-n\\0&1\end{pmatrix}.$$

From this, can you describe the subgroup generated by $a$?

As noted below in the comments, we have an isomorphism $$\langle\begin{pmatrix}1&1\\0&1\end{pmatrix}\rangle\cong\mathbb Z,$$ given by sending your generator $a$ to the element $1\in\mathbb Z$. But to get a description of $\langle a\rangle$ as a subgroup of the $2\times 2$ complex matrices, we noted that the above facts imply that $$\langle a\rangle=\left\{\begin{pmatrix}1&n\\0&1\end{pmatrix}:n\in\mathbb Z\right\}.$$

Now, to make sure you understand all of this, convince yourself that given a complex number $\mu\in\mathbb C$, we have the following description: $$\langle \begin{pmatrix}1&\mu\\0&1\end{pmatrix}\rangle=\left\{\begin{pmatrix}1&n\mu\\0&1\end{pmatrix}:n\in\mathbb Z\right\}.$$

• It seems that $a$ is an infinite group. It seems $a*a$ gives a new element, call it $t$, that needs a new inverse element. And then $t*t$ will give a new element...and so on...So the smallest group is an infinity large group? – dukenukem Oct 2 '12 at 19:56
• @dukenukem You're correct that it's an infinite group, but I doubt that's enough information for a complete answer since there are tons of infinite groups of matrices. What elements are in this group? Is it isomorphic to any well-known infinite groups you already know about? – MartianInvader Oct 2 '12 at 20:27
• With MTurgeon's calculations it should be more or less evident, that your group is isomorphic to $\mathbb Z$ via $\left(\begin{matrix}1&n\\0&1\end{matrix}\right)\mapsto n$. – Hagen von Eitzen Oct 2 '12 at 20:28
• Ok I can see that now...The group can be mapped to $\mathbb{Z}$. So how do I put my answer to the question. This is the problem I am really having with group theory...I don't know how to formulate answers. I am asked to describe the smallest group of 2x2 complex matrices containing $a$..but even though I know what this group is and can see that it is isomorphic to $\mathbb{Z}$ I don't know how to 'describe the smallest group of 2x2 complex matrices containing $a$'? Do I just say 'the infinite group generated by $\begin{pmatrix}1&n\\0&1\end{pmatrix}$'? – dukenukem Oct 2 '12 at 20:43
• @dukenukem see the edit. – M Turgeon Oct 2 '12 at 23:23