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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?

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    Ok ty, so the smallest possible group containing $a$ is infinitely large?2012-10-02

1 Answers 1

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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: $\langle \begin{pmatrix}1&\mu\\0&1\end{pmatrix}\rangle=\left\{\begin{pmatrix}1&n\mu\\0&1\end{pmatrix}:n\in\mathbb Z\right\}.$

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    @dukenukem see the edit.2012-10-02