Originally Posted by

**ChrisBickle** Here is the question: Let

H = 2x2 matrix with entries

|cosA -sinA |

|sinA cosA|

and prove H is a subgroup of SL2(R)

Showing the inverse is an element isnt a problem im having a problem showing closure(havent attempted inverse yet)

I let A,B element of H then u get 2 matrices with those entries and A and B as measures then i get

AB = |cosAcosB - sinAsinB -cosAsinB -sinAcosB|

|sinAcosB + cosAsinB -sinAsinB + cosAcosB|

then det(AB) =

(cosAcosB - sinAsinB)(-sinAsinB + cosAcosB) - (-cosAsinB - sinAcosB)(sinAcosB + cosAsinB)

when you multiply it all out i get abunch of cosAcosBsinAsinB terms that cancel off a cos(squared)Acos(squared)B and sin(squared)Asin(squared)B that add to 1 and then i get 2 other terms a cos(squared)Asin(squared)B and sin(squared)Acos(squared)B. I know that the det = 1 is there some identity with the other 2 terms im missing or did i make a mistake somewhere else. Thanks in advance for your help.