# Order a = Order a inverse proof i have no idea how to start

• Oct 26th 2009, 11:00 PM
ChrisBickle
Order a = Order a inverse proof i have no idea how to start
Hey everyone im taking a first semester set theory class and im trying to get ahead on hw since im not doing well and i was wondering if anyone could give me some kind of hint or sumthin to start this proof for hw:

Let G be a group, Prove ord(a) = ord(a)inverse the hint for the problem is to prove 2 cases G in finite order and G is infinte order. We have gone over this kinda stuff im just not sure where to start. If G is finite order then for some integer m a^m = e show that a^m(inverse) = e? thus order(ainverse) = order(a) and for infinite you would show for infinite order for a, ainverse has infinite order?
• Oct 27th 2009, 05:03 AM
tonio
Quote:

Originally Posted by ChrisBickle
Hey everyone im taking a first semester set theory class and im trying to get ahead on hw since im not doing well and i was wondering if anyone could give me some kind of hint or sumthin to start this proof for hw:

Let G be a group, Prove ord(a) = ord(a)inverse the hint for the problem is to prove 2 cases G in finite order and G is infinte order. We have gone over this kinda stuff im just not sure where to start. If G is finite order then for some integer m a^m = e show that a^m(inverse) = e? thus order(ainverse) = order(a) and for infinite you would show for infinite order for a, ainverse has infinite order?

In groups it's true the same as with real numbers: $\left(g^{-1}\right)^k = g^{-k}=\left(g^k\right)^{-1}$

Tonio