# Thread: order of a group

1. ## order of a group

My Question is :
Let G=(a b , c d) such that a,b,c,d belongs to Z(3)={0,1,2} and
det(G)=ad-bc not equal to 0
Show that order of G=48
In fact I know that order of a finite group is equal to order of generator of it. i.e if G is cyclic group generated by a, then the order of a is same as the order of G, namely the smallest positive integer such that a^k=e
But it seems that this will not be helpful here. How should I start in order to proof that this group is of order 48

Thank you in advance

2. Originally Posted by fuzzy topology
My Question is :
Let G=(a b , c d) such that a,b,c,d belongs to Z(3)={0,1,2} and
det(G)=ad-bc not equal to 0
Show that order of G=48
In fact I know that order of a finite group is equal to order of generator of it. i.e if G is cyclic group generated by a, then the order of a is same as the order of G, namely the smallest positive integer such that a^k=e
But it seems that this will not be helpful here. How should I start in order to proof that this group is of order 48
There's probably a slick group-theoretic way of doing this, but I would just use a simple counting strategy.

There are $81=3^4$ elements in $\mathbb{Z}(3)^4$, which I'm thinking of as the set of all 2×2 matrices over $\mathbb{Z}(3)$. Count the number of those for which the determinant is equal to 0:
there is one matrix with all four components zero (the zero element),
there are 8 matrices with three components zero,
there are 8 matrices with two zero components (in the same row),
there are 8 matrices with two zero components (in the same column).
There are no matrices with zero determinant and just one zero. Finally, there are 8 matrices with zero determinant and no zero components (a, b and c can all be 1 or 2, and then d is determined by the condition ad–bc=0).

That gives a total of 33 matrices with determinant 0, hence 81 – 33 = 48 matrices with nonzero determinant.

3. Thank you very much> I tried to solve it by other ways but I did not succeed. Thanks alot