Thread: Proving a subgroup relation

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.

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.

this seems to be basic trigonometry: for ex. cosAcosB - sinAsinB = cos(A+B)...check your trigonometric identities.

Tonio

Thank you for your help.

Chris