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.

My answer:

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?

  • 1
    $\begingroup$ 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. $\endgroup$ – anon Oct 2 '12 at 19:46
  • $\begingroup$ What does <A> mean? I am trying to understand the process for answering these questions. $\endgroup$ – dukenukem Oct 2 '12 at 19:48
  • 1
    $\begingroup$ 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. $\endgroup$ – anon Oct 2 '12 at 19:49
  • $\begingroup$ Ok ty, so the smallest possible group containing $a$ is infinitely large? $\endgroup$ – 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: $$\langle \begin{pmatrix}1&\mu\\0&1\end{pmatrix}\rangle=\left\{\begin{pmatrix}1&n\mu\\0&1\end{pmatrix}:n\in\mathbb Z\right\}.$$

  • $\begingroup$ 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? $\endgroup$ – dukenukem Oct 2 '12 at 19:56
  • 1
    $\begingroup$ @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? $\endgroup$ – MartianInvader Oct 2 '12 at 20:27
  • $\begingroup$ With MTurgeon's calculations it should be more or less evident, that your group is isomorphic to $\mathbb Z$ via $\left(\begin{matrix}1&n\\0&1\end{matrix}\right)\mapsto n$. $\endgroup$ – Hagen von Eitzen Oct 2 '12 at 20:28
  • $\begingroup$ 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 $\begin{pmatrix}1&n\\0&1\end{pmatrix}$'? $\endgroup$ – dukenukem Oct 2 '12 at 20:43
  • 1
    $\begingroup$ @dukenukem see the edit. $\endgroup$ – M Turgeon Oct 2 '12 at 23:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.