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Show that $$G=\left\{\begin{pmatrix}a\ b \\ c\ d\end{pmatrix} : a,b,c,d\in\mathbb{Z}_3, ad-bc\ne 0 \text{ in }\mathbb{Z}_3\right\}$$ is a group with the usual matrix multiplication and $|G| = 48$.

I already proved that it's a group, but I'm bit lost on how to "efficiently" determine the order of such a group (that is, not just listing and counting, I also say this because I'll need some way to determine the cardinality of some larger groups).

At first I looked for some theorem on groups that I've already seen but found none of them to be useful here (surely I must be wrong on this). Then I thought that there are $3^4 = 81$ matrices with entries on $\mathbb{Z}_3$ and then I need to exclude those with $ad=bc$. Any hints to continue on this line? Or for calculating order of groups in general?

Thanks in advance.

Shaun
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AnalyticHarmony
  • 2,085

2 Answers2

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Hint:

It's quite simple:

  • first you choose the non-zero column vector $\begin{pmatrix}a\\c\end{pmatrix}$ in $\mathbf F_3^2$, which makes $mmm$ possibilities,
  • then you choose the second column vector $\begin{pmatrix}b\\d\end{pmatrix}$ , which must not be a scalar multiple of the previous choice, which eliminates $nnn$ vectors.
Bernard
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In general, $\operatorname{GL}(n,\Bbb Z_p)$ has order $(p^n-1)(p^n-p)\cdots(p^n-p^{n-1})$.

In this case, we get $(3^2-1)(3^2-3)=8(6)=48$.