# Is there something wrong with this question concerning Groups

We consider the group $$G = SL(2, 3)$$ i.e, the set of $$2 \times2$$ matrices with determinant 1 and addition and multiplication are performed modulo 3 even in the determinant formula. One can show that $$|G| = 24$$

a) Let $$\alpha = \begin{pmatrix} 2 & 2\\ 2 & 1 \end{pmatrix}$$ show that $$\alpha \in G$$ and find its inverse $$\alpha^{-1}$$.

b) Let H = {$$\begin{pmatrix} a & b\\ 0 & a \end{pmatrix}$$ where a, b $$\in$$ {0,1,2}, a $$\neq 0$$}. Show that H is a subgroup and find a familiar group that is Isomorphic to H.

c) The subgroup H contains two elements of order 3. Find 8 elements of order 3 in G.

HINT: H only contains upper triangular matrices; also use conjugation.

It's trivial to show H $$\leq$$ G and I found it's Isomorphic to $$D_{6}$$. It's the last part I have found issues in my understanding.

I wrote out all the elements of H:

$$\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$$ $$\begin{pmatrix} 2 & 0\\0 & 2 \end{pmatrix}$$ $$\begin{pmatrix} 2 & 1\\0 & 2 \end{pmatrix}$$ $$\begin{pmatrix} 2 & 2\\0 & 2 \end{pmatrix}$$ $$\begin{pmatrix} 1 & 1\\0 & 1 \end{pmatrix}$$ $$\begin{pmatrix} 1 & 2\\0 & 1 \end{pmatrix}$$

Four of these have an order of 3. Not just two of them. Or is my understanding off?

• You said that $H$ is isomorphic to $D_6$ (the group of symmetries on a triangle). The elements $r$ and $r^2$ have order 3 in $D_6$. Which elements in $H$ correspond to those elements? $D_6$ does not have 4 elements that have order 3 as $f$, $rf$, and $r^2f$ have order 2 and $e$ has order 1. Dec 3 '21 at 1:45
• @Isaiah I used a theorem to come to that conclusion. $|H| = 6$ and 6 = 2p where p is a prime number number greater than 2, that is p=3. Then the group H is isomorphic to either $Z_{6}$ or $D_{6}$. Since H is non-abelian then it must be the case H is isomorphic to the latter. Dec 3 '21 at 1:56

I checked the matrices in your subgroup $$H$$, and I found that two of them had order 6, two of them had order 3, one had order 2, and one had order 1. This tells me that $$H$$ is not isomorphic to $$D_6$$ because $$D_6$$ does not have any elements of order 6.