Consider the group $G = GL(2, \Bbb Z_{11})$. How many elements does $G$ have?

Question

Consider the group $G = GL(2,\Bbb Z_{11})$. How many elements does $G$ have? Any ideas as to how I would go about finding it?

can I make permutations?. I do not have an idea yet. Help, please.

Think how many choices of $(a,b,c,d)$ yield $ad-bc=0$ where $a,b,c,d$ are coming from $\mathbb Z_{11}$. — Arkady, Feb 01 '18 at 05:34
Think of how many nonzero 1st rows are possible. Then think of how many 2nd rows are linearly independent of any given 1st row. — Gerry Myerson, Feb 01 '18 at 06:08
Possible duplicate of Order of general- and special linear groups over finite fields. — ahulpke, Feb 01 '18 at 20:07

score 7 · Accepted Answer · answered Feb 01 '18 at 06:58

First we consider the first row. First row is a non-zero vector, we have $11^{2}-1=120$ choices. Then we think about the second row. The second row vector and the first one are linearly independent, so we have $11^{2}-11=110$ choices. Hence, there are $120\times110=13200$ elements in $\mathrm{GL}(2,\mathbb{Z}_{11}).$

In general, we have the formula:$|\mathrm{GL}(n,\mathbb{F}_{q})|=\prod\limits_{j=0}^{n-1}(q^{n}-q^{j}).$

Consider the group $G = GL(2, \Bbb Z_{11})$. How many elements does $G$ have?

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