Does Proof by Contradiction Confirm the Order of Elements in a Group?

[annoymage]

lal means the order of a

Theorem.

Let G be a group and a$$\in$$G. Then lal=la^-1l

Proof.

Case 1, if lal=1

Case 2, if lal=n

Case 3, if lal=infinity

i understand case 1 and 3, so i'll be post the proof when need,

but, case 2

Here's the proof

Suppose lal=n

Then aⁿ=1 and a^r$$\neq$$1, 1$$\leq$$r<n-----------(1)

To show that (a^-1)ⁿ=1 and (a^-1)^r$$\neq$$1, 1$$\leq$$r<n.

Clearly, (a^-1)ⁿ=(aⁿ)^-1=(1)^-1=1

Suppose (a^-1)^r=1, for some 1$$\leq$$r<n

=> (a^r)^-1, 1$$\leq$$r<n

=> a^r=1, for some 1$$\leq$$r<n

but this contradict (1)

So, (a^-1)^r$$\neq$$1, 1$$\leq$$r<n

Hence, lal=n=la^-1l
I don't understand why, it suppose "(a^-1)^r=1, for some 1$$\leq$$r<n"

then say it contradict with (1), i cannot see how they contradict.

Help, T_T

 

[Office_Shredder]

=> (a^r)^-1, $$1 \leq r<n$$

You need to finish this line by having an equation, not an expression. I suspect you meant to put =1 in there.

You showed that if the order of a^-1 is r with r<n, then a^r=1. But the order of a was n, so that's the contradiction

 

[losiu99]

(a^-1)^r=a^-r=a^n-r, so if 0<r<n and a^-r=e, order of a is not n.

 

[annoymage]

You need to finish this line by having an equation, not an expression. I suspect you meant to put =1 in there.

You showed that if the order of a^-1 is r with r<n, then a^r=1. But the order of a was n, so that's the contradiction

OOOOOOOOOOOOOOOOOOOOO, i get it, now I'm trying to catch what losiu99 try to convey.
in the mean time

do you mind checking this? please :D

https://www.physicsforums.com/showthread.php?t=417859

 

[losiu99]

Sorry, my post was a bit off topic, I thought you didn't understand the part on reaching a^r=1 starting from (a^-1)^r.