# show that the group of $2\times 2$ invertible matrices is isomorphic to $S_{3}$

1) Show that the set of 2x2 matrices with non zero determinat together with multiplication form a group $(H,*)$ where $$H=\left.\left\{\begin{pmatrix} a&b\\c&d\end{pmatrix}\right|\; a,b,c,d\in \mathbb{Z}_2, ad-bc\neq 0 \right\}$$ Show that this group is isomorphic to $S_{3}$

2) identify elements of order 2 and show they are characterized by their traces

Hint show that the group acts naturally on the vector space $\mathbb{Z_{2}}\oplus \mathbb{Z_{2}}$ and permutes the non-zero elements.

part 1 is very simple. I showed that the multiplications is an operation on this set, the associativity is inherited from the 2x2 matrices, the id element is just the id matrix and the set is closed under taking inverses.

How to do the part 2)? How to create such isomorphism? What does it mean that the elements of order 2 are characterized by their traces???

Hint: Let $H$ act canonically on the set of non-zero vectors of $\Bbb Z_2\oplus \Bbb Z_2$ and show that this action is faithful. Then $H$ must be isomorphic to a subgroup of $S_3$ and simply by counting it must be all of $S_3$.
"The elements of order $2$ are characterised by their trace" means that there is a condition of the form "$A\in H$ has order $2$ if and only if $\text{trace}A=\;...$
Edit: Think of the non-zero vectors $\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}$ and $\begin{pmatrix}1\\1\end{pmatrix}$ as the numbers $1, 2, 3$. Formally, we take a bijection (a "renaming" if you will) $f$ from the set of non-zero vectors into the set $\{1,2,3\}$. Now, regular matrices map non-zero vectors onto non-zero vectors, so we can define a map $\varphi:GL(2,\Bbb Z_2)\to S_3$ by the rule $A\mapsto \sigma$, where $\sigma(i)=f(Af^{-1}(i))$. That means that if, for example, $A\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}0\\1\end{pmatrix}$, then the corresponding $\sigma$ will satisfy $\sigma(1)=2$. This $\varphi$ will give you the desired isomorphism.
• It is just very new to me. Our previous teacher taught us that to show isomorphism we need to find a bijective function that is a homomorphism. In the examples like the one above I have noticed that one usually just shows e.q that the order of two groups is the same. I found that the order of the group H must be 6, since we have 24 matrices having entries in $Z_{2}$ but only 6 of them have non-zero determinant. $S_{3}$ have also order 6. Both groups are non-commutative...but is it enough to prove that there exist an isomorphism between them? "A∈H has order 2 if and only if traceA=...2???'' – H.E Sep 8 '13 at 12:53
• one question more is my group H the same thing as $GL(2,\mathbb{Z_{2}})$? – H.E Sep 8 '13 at 13:02
• Yes the group $H$ is the same thing as $GL(2,\Bbb Z_2)$, because the non-singular matrices (non-zero determinant) are precisely the regular (invertible) matrices. Since $H$ is isomorphic to $S_3$, $H$ has precisely $3$ elements of order $2$. Now, the trace of an $n\times n$-matrix is $-1$ times the coefficient of $x^{n-1}$ in the characteristic polynomial. If $A$ has order $2$, then $A^2=I$ so its characteristic polynomial is $x^2-1$, so the trace is $0$. (Note $0=2$ in $\Bbb Z_2$.) – walcher Sep 8 '13 at 13:56