# Matrix representation of $\operatorname{Aut}(\mathbb{Z}_2\times\mathbb{Z}_2)\cong\mathbb{S}_3$ [duplicate]

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.

## marked as duplicate by Matthew Leingang, David Hill, E. Joseph, Namaste abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 17 '16 at 0:21

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.
• 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? – Allen Ai Nov 16 '16 at 16:03
• Typo: I meant $\mathbb Z_2^2. \qquad$ – Michael Hardy Nov 16 '16 at 16:21