My Math Forum order in GL(2,R)

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 October 10th, 2007, 08:10 PM #1 Newbie   Joined: Oct 2007 Posts: 7 Thanks: 0 order in GL(2,R) In GL(2,R), let P= (0 -1) (1 1) Q= (0 1) (-1 1). a)compute to show that P has order 6 and Q has order 4. (and check that the orders cannot be smaller) b)compute PQ, prove by induction what (PQ)^n is, and show that PQ has infinite order.
 October 16th, 2007, 08:47 PM #2 Member   Joined: Mar 2007 Posts: 57 Thanks: 0 I assume by "order" you mean the lowest power you have to raise to get an element to the identity? For part a) Compute P^2, P^3, P^4, P^5, P^6 and hopefully you get P^6 = identity in GL_2 (R) while the others are not. Similarly you can compute Q^2, Q^3, Q^4 and hope you get Q^4 = identity while Q^2 and Q^3 are not. For part b) compute PQ = (1 -1) (-1 1) (PQ)^2 = PQPQ = (2 -2) (-2 2) So you should be able to use induction to prove for all n, (PQ)^n= (n -n) (-n n) Hence (PQ)^n cannot be the identity in GL_2 (R) for any finite n, which means (PQ) does not have finite order.

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### order of gl2r

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