9

I need to show that this set is closed under matrix multiplication, is there a better way than doing it via a Cayley table? Or rather I assume there is and I just can't get my head around it. Any help would be greatly appreciated.

$\begin{gather*} M= \left\{ \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & -1\\ -1 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}, \begin{bmatrix} -1 & 0\\ 0 & -1 \end{bmatrix}, \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}, \begin{bmatrix} -1 & 0\\ 0 & 1 \end{bmatrix} \right\}\\ \end{gather*}$

ifearthenight
  • 161

8 Answers8

21

Call matrices of this $\begin{pmatrix}*&0\\0&*\end{pmatrix}$ form "diagonal" and matrices of this $\begin{pmatrix}0&*\\*&0\end{pmatrix}$ form "skew diagonal". First prove that the product of diagonal matrices is diagonal, the product of diagonal and skew diagonal is skew diagonal, the product of skew diagonal and diagonal is skew diagonal and that the product of skew diagonal and skew diagonal is diagonal.

Then prove that every entry in the resulting matrices will always be $\pm 1$. Finally note that every matrix which is diagonal or skew diagonal and has entries in $\{\pm 1\}$ is one of your eight.

Oscar Cunningham
  • 16,939
9

The matrices $A = \begin{bmatrix}1&0\\0&-1\end{bmatrix}$ and $B = \begin{bmatrix}0&1\\1&0\end{bmatrix}$ satisfy the identities $BB=I$, $AA = I$, $AB=-BA.$ Consider the group of all possible matrix products of $A$s and $B$s, consider any element of it. The last identity allows you to reorder any matrix multiplication of these two into the form $\pm A\dots AB\dots B$ and the first two identities allow you to remove As and Bs from it two-at-a-time, so the group has at most 8 elements, $$\{I, A, B, AB, -I, -A, -B, -AB\}.$$ By inspection, these are the 8 that you have and none of them is secretly equal to another of them, so the group has order 8.

CR Drost
  • 3,027
5

Here's a second way to do it which might be quicker. Define $G$ to be the group of the $2\times 2$ invertible matrices which fix the set $\{(0,1),(1,0),(0,-1),(-1,0)\}$. It's easy to prove this is a group and also easy to check that it contains precisely your eight matrices.

Oscar Cunningham
  • 16,939
3

A first step is to determine the order of the $8$ elements. We can easily see that not all matrices commute, so that we have two distinct non-abelian groups of order $8$, which may be represented by these matrices, namely the dihedral group $D_4$ or the quaternion group $Q_8$. Now the information of the order of elements is helpful, because we see that it represents the dihedral group - which is closed under multiplication. Also, here we see that this is correct.

Dietrich Burde
  • 140,055
2

One way would be to demonstrate a group isomorphism between your set and a group where you know the operation is closed.

If you're familiar with the dihedral group of order 8, that might be a good place to start. Then note that as the dihedral group is generated by the basic operations of rotations and flips, so it would be sufficient to show that the group isomorphism works for those two elements mapped to $R$ and $F$.

kholli
  • 464
2

Your group $M$ can be described

$$M = \left\lbrace \begin{pmatrix} a_1 &0\\ 0 &b_1\end{pmatrix}, \begin{pmatrix} 0 &a_2\\ b_2 &0\end{pmatrix} : a_i, b_i \in (\Bbb Z/3\Bbb Z)^\times\right\rbrace.$$

Then you need only note that any product $a_1a_2$ or $b_1b_2$ lands back in $(\Bbb Z/3\Bbb Z)^\times$ as this is a group.

Edward Evans
  • 4,739
  • 1
    The trouble with this approach is that you don't know a priori that you won't get, say, $$\begin{pmatrix} 7 & 0\ 0 & 3\end{pmatrix}$$ if you multiply the original matrices. Of course, it is easy to see that this is impossible, but you'd have to basically include what Oscar Cunningham wrote, or something similar. – Ennar Apr 14 '17 at 16:12
  • 1
    Assuming that the original post meant to describe a group of integer matrices, what you describe isn't literally correct. A collection of matrices over $\mathbb Z$ is not the same as a collection of matrices over $\mathbb Z/3\mathbb Z$, even if they look the same and have the same multiplication table (which, as @Ennar points out, needs to be checked anyway). One should say instead that they are isomorphic, or, if desired, emphasise the 'canonicity' of the isomorphism by saying that the natural map of matrices induced by $\mathbb Z \to \mathbb Z/3\mathbb Z$ induces an iso. onto its image. – LSpice Apr 14 '17 at 18:33
2

This is not a rigorous proof, but an interesting geometric intuition. Those matrices are different kinds of reflections (about origin, about $y=x$, etc). As an example, pick $(2,1)$ and transform it with each of those matrix. You have (label $i$ means the $i$th matrix is applied)

reflection

It's apparent that those transformations are "geometrically closed".

Taozi
  • 2,743
0

Another geometric way: Fix the standard orthonormal frame. Then the given diagonal matrices each either fix or reflect the basis vectors, e.g. $\left(\begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix}\right)$ fixes the first and reflects the second. The matrix $\left(\begin{smallmatrix} 0 & 1 \\ 1 & 0\end{smallmatrix}\right)$ is a change of orientation and all skew diagonal ones are given by this change of orientation, e.g. $\left(\begin{smallmatrix} 0 & 1 \\ -1 & 0\end{smallmatrix}\right)=\left(\begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix}\right)\left(\begin{smallmatrix} 0 & 1 \\ 1 & 0\end{smallmatrix}\right)$. We may assign a triple $(a,b,c)$ from $\{-1,1\}\times \{-1,1\}\times \{-1,1\}$ to each matrix, where $a$ indicates whether the first basis vector is reflected or not, $b$ for the second and $c$ tracks whether we had a change of orientation. Hence we see that we can identify your matrices with the group $\mathbb{Z}_2^3$ which is closed.

Sebastian Bechtel
  • 2,153