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Show that the group $\mathrm{GL}_2(\mathbb{F}_5)$ has order 480. By defining a suitable homomorphism from $\mathrm{GL}_2(\mathbb{F}_5)$ to another group, which should be specified, show that the order of $\mathrm{SL}_2(\mathbb{F}_5)$ is 120. Find a subgroup of $\mathrm{GL}_2(\mathbb{F}_5)$ of index 2.

For the first part I actually can show the order 480. First I find the order of $\mathrm{M}_2(\mathbb{F}_5)$ , $5^4=625$, then I discard all the singular matrices then I have 480. But find all the cases that the matrices are singular is tedious and not easy to ensure every case is considered in the field. I want to find a simpler and more effective method to approach this.

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Hint: how many ordered basis has a $\,2-$dimensional vector space over $\,\Bbb F_5:=\Bbb Z/5\Bbb Z\,$ ? – DonAntonio Dec 30 '12 at 3:22
2… This answer contains an explanation, but don't divide by n! ... – Jason Polak Dec 30 '12 at 3:48

The group acts transitively on the set of non-zero vectors in the two-dimensional vector space $(F_5)^2$, which has $5*5 - 1 = 24$ elements. Can you calculate the stabilizer, for say, the vector $(1,0)$? Don's approach works but it probably involves the same amount of computations as you've already done.

Finally, the determinant map is your friend.

I would have left this as a comment but I don't have the reputation. And I could just spell out the answer if you want, but I figured these hints would be enough.

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About Don's approach: really? $24\cdot (24-4)=480$. Not that much calculation. – tomasz Dec 30 '12 at 3:40
Or, to highlight the general rule, $(25-1)\cdot(25-5)=480$. – tomasz Dec 30 '12 at 3:50
Ah, tomasz, you're obviously right. My mistake. – user54535 Dec 30 '12 at 4:59

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