1
$\begingroup$

We can show that $\operatorname{Aut}(\mathbb{Z}_{2}\times\mathbb{Z}_2)\cong\mathbb{S}_3$ in two ways:
(a). One way is to show that any automorphism is determined by an invertible 2-by-2 matrix with entries in $\mathbb{Z}_2$, that there are six such matrices, and that they form a group isomorphic to $S_3$.

I don't quite understand how to show that any automorphism is determined by such 2-by-2 matrix, e.g. what matrix are we looking for and how do they determine the automorphism. Can someone please enlighten me with, maybe, some concrete examples? Thanks!

Edit: This is not a duplicate as I'm not looking for a proof but rather seeking examples of this particular way of looking at the problem.

$\endgroup$

marked as duplicate by Matthew Leingang, David Hill, E. Joseph, Namaste abstract-algebra Nov 17 '16 at 0:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    $\begingroup$ See math.stackexchange.com/questions/693646 $\endgroup$ – Watson Nov 16 '16 at 15:40
  • $\begingroup$ @Watson I see. But I still seek a concrete example with this matrix approach though! $\endgroup$ – Allen Ai Nov 16 '16 at 15:42
  • $\begingroup$ It is exactly like a 2 times 2 invertible real matrix gives an automorphism of the real plane. $\endgroup$ – quid Nov 16 '16 at 15:44
  • $\begingroup$ @MatthewLeingang The post you mentioned asked for the proof of a particular statement, while I am seeking some concrete examples of a way to represent the said automorphism with matrix. I have read the linked post yet the examples offered in that post does not give me enough information to work on the question. Hence I don't consider this post a duplicate. Thank you. $\endgroup$ – Allen Ai Nov 16 '16 at 15:57
1
$\begingroup$

The three nonzero members of $\mathbb Z_2^2$ are $$ a=\left[ \begin{array}{r} 1 \\ 0 \end{array} \right], \quad b= \left[ \begin{array}{r} 0 \\ 1 \end{array} \right], \quad c=\left[ \begin{array}{r} 1 \\ 1 \end{array} \right]. $$ So you want to represent the six permutations of $\{a,b,c\}$ by matrices. So ask yourself, for example, for which matrix $\left[ \begin{array}{cc} k & \ell \\ m & n \end{array} \right]$ do you have $$ \left[ \begin{array}{cc} k & \ell \\ m & n \end{array} \right] \left[ \begin{array}{r} 1 \\ 0 \end{array} \right] = \left[ \begin{array}{r} 1 \\ 1 \end{array} \right] \text{ and } \left[ \begin{array}{cc} k & \ell \\ m & n \end{array} \right]\left[ \begin{array}{r} 0 \\ 1 \end{array} \right] = \left[ \begin{array}{r} 0 \\ 1 \end{array} \right] $$ and so on.

$\endgroup$
  • $\begingroup$ Hmm. It seems like I have misinterpreted the statement. I thought $\mathbb{Z}_2$ is the cyclic group of order 2, hence it's different from $\mathbb{Z}^2$. Could you please help me clarify it? $\endgroup$ – Allen Ai Nov 16 '16 at 16:03
  • $\begingroup$ Typo: I meant $\mathbb Z_2^2. \qquad$ $\endgroup$ – Michael Hardy Nov 16 '16 at 16:21

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