This time, it's about proving 2 elements have the same order:-

Let G be a group and x be an element of G. The$\displaystyle x^n$orderof x is the least positive number such that= e.. That is, n$\displaystyle x^n$>1 satisfies

= e and$\displaystyle x^m$=/= e for all 1.<m<n

For x in G prove that x and x^{-1}have the same order

So, I've had a fair whack at the question, but I'm not sure if I'm going in the write direction.

$\displaystyle x^{n/2}$.$\displaystyle x^{n/2}$ = $\displaystyle x^n$= e

$\displaystyle x.x^{-1}$ = e

$\displaystyle x.x^{-1}$ = $\displaystyle x.x^-1$

= $\displaystyle x^{1-1}$

= $\displaystyle x^0$

=1 ---> e (identity is 1)

$\displaystyle x^n$ = e

$\displaystyle (x^{n})^{-1} = e^{-1}$

$\displaystyle x^{-n} = e^{-1}$

Note: $\displaystyle e^{-1}$ = $\displaystyle e$

Therefore,

$\displaystyle x^{-n} = e = x^{n}$

$\displaystyle x^{-n} = x^{n}$

$\displaystyle x^{n}/x^{-n} = e$

$\displaystyle x^{n}.x^{n} = e$

$\displaystyle x^{2n} = 1 = e$

$\displaystyle e = 1 = x^{0}= x^{-n}= x^{n}= x^{2n}$

THUS. If

$\displaystyle x^{n}$ = 1 = e

$\displaystyle x^{n}.x^{-n}$ = 1 = e

$\displaystyle x^{-n}$ = 1 = e

$\displaystyle x^{n}$ = $\displaystyle x^{-n}$

Therefore they have the same order.

That is huge im sorry, but yeh. I dunno if it's correct. Any pointers?