# Homework Help: Proof by contradicton

1. Jul 26, 2010

### 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 ar$$\neq$$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

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

=> ar=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

2. Jul 26, 2010

### Office_Shredder

Staff Emeritus
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 ar=1. But the order of a was n, so that's the contradiction

3. Jul 26, 2010

### losiu99

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

4. Jul 26, 2010

### annoymage

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