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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?

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    $\begingroup$ 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. $\endgroup$
    – Isaiah
    Dec 3 '21 at 1:45
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    $\begingroup$ @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. $\endgroup$
    – Malcolm
    Dec 3 '21 at 1:56
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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.

You claim that this subgroup is not Abelian. While this is true for matrices in general, certain special matrices do in fact commute. All the matrices in this subgroup commute, so the subgroup is Abelian.

I would advise performing the multiplications again and remember to mod the entries by 3. It is easy to make an error.

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  • $\begingroup$ Thank you. This helps a lot. $\endgroup$
    – Malcolm
    Dec 3 '21 at 2:23

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