Set is closed as for multiplication of matrices We have the matrices \begin{equation*}s:=\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}, \ d:=\frac{1}{2}\begin{pmatrix}-1 & -\sqrt{3} \\ \sqrt{3} & -1\end{pmatrix}\end{equation*} and the points \begin{equation*}p:=\begin{pmatrix}2 \\ 0 \end{pmatrix}, \ q:=\begin{pmatrix}-1 \\ \sqrt{3} \end{pmatrix}, \ r:=\begin{pmatrix}-1 \\ -\sqrt{3} \end{pmatrix}\end{equation*}
I draw the points $p, q, r$ and calculate also the points $sp, sq, sr, dp, dq, dr$ and I noticed that $s$ is a reflection as for the $x$-axis and $d$ is a rotation of $\frac{2\pi}{3}$.

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*Consider $G:=\{d, d^2, d^3, sd, sd^2, sd^3\}\subseteq \mathbb{R}^{2\times 2}$ and show that $G$ is closed as for multiplication of matrices, i.e. $gh\in G$ for all $g,h\in G$.


*Show that the elements of $G$ are invertible and for each $g\in G$ there is $g^{-1}\in G$.


*What is the geometric interpretation of $G$ ?


*Let $z=\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}$ and $H:=G\cup \{zg\mid g\in G\}$. Show that $H\subseteq \mathbb{R}^2$ is closed as for multiplication of matrices.
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Let's start with question 1. Do we have to consider all possible combinations of the elements of $G$ and show that their product is again in $G$ ?
For example, do we have to do the following?
\begin{align*}d\cdot d^2&=d^3\in G \\ d\cdot d^3&=d^4=4-\text{times rotation about }120^{\circ}=\text{ratotaion about }480^{\circ}\\ & =\text{rotation about }360^{\circ}+120^{\circ}=\text{rotation about }120^{\circ}\\ & =d\in G \\ d\cdot s\cdot d&=\text{rotation about }120^{\circ}\text{ then reflection about x-axis and then rotation about }120^{\circ}\\ & =\text{reflection abour ax-axis}=s\in G\end{align*}
Or is there is shorter way?
At question 2 do we have to show that the matrices are invertible? Or do we have to consider what the matrices represent?
 A: There are several useful relations you can notice.

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*Since three rotations by $120^\circ$ give rotation by $360^\circ$, we have $d^3 = I.$

*Rotating a point by $0^\circ$, $120^\circ$ and $240^\circ$ gives an equilateral triangle around the origin so the sum of those points is $0$. Therefore $d^2+d+I=0$.

*Reflecting around $x$-axis twice doesn't change anything so $s^2=I$.

*Algebraically, multiplying a matrix by $s$ multiplies a row or a column of a matrix by $-1$ so we get $$sd+ds = \frac12 \begin{bmatrix} -1 & \sqrt{3} \\ -\sqrt{3} & 1 \end{bmatrix}+\frac12 \begin{bmatrix} -1 & -\sqrt{3} \\ -\sqrt{3} & 1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} = -s$$
and therefore $$ds = -sd-s = -s(d+I)=-s(-d^2) = sd^2.$$

*Similarly, $$sd = sd^4 = (sd^2)d^2 = (ds)d^2 = d(sd^2) = d(ds) = d^2s.$$
Therefore, your set $G$ is $$G = \{I,d,d^2,s,sd,sd^2\} = \{I,d,d^2,s,ds,d^2s\}.$$
Denote $S = \{I,s\}$ and $D = \{I,d,d^2\}$ and it is easy to see that $S$ and $D$ are closed under multiplication. Now the crucial observation is that $G$ is the element-wise product of $S$ and $D$ in both orders:
$$G = SD = DS.$$
Take two elements $x,y \in G$. We can write $x = s_1d_1$ and $y = d_2s_2$ with $s_1,s_2 \in S, d_1,d_2 \in G$ so
$$xy = (s_1d_1)(d_2s_2) = s_1(d_1d_2)s_2 = s_2d_3s_2$$
for some $d_3 \in D$ since $D$ is closed under multiplication. Now $d_3 s_2 = s_2'd_3'$ for some $s_2' \in S$ and $d_3' \in D$ so
$$xy = s_2(d_3s_2) = s_2s_2'd_3' = s_3d_3' \in SD = G$$
for some $s_3 \in S$ since $S$ is closed under multiplication. Therefore $xy \in G$ so $G$ is closed under multiplication.
It is also very evident that all matrices in $G$ are invertible. Namely, $s$ and $d$ are invertible with inverses $s^{-1}=s$ and $d^{-1}=d^2$ and hence all matrices in $S$ and $D$ are invertible as products of invertible matrices. Moreover, element-wise we have $S^{-1} = S$ and $D^{-1}=D$. Now for $x \in G$ we have $x = s_1d_1$ for some $s_1\in S$, $d_1 \in D$. We have
$$x^{-1} = (s_1d_1)^{-1}=d_1^{-1}s_1^{-1} \in D^{-1}S^{-1}=DS = G$$
so $x^{-1} \in G$.
The relevant reference is the result that element-wise product $EF$ of two subgroups $E, F$ is itself a subgroup if and only if $EF = FE$.
