# Proof about orders of elements. Ord(a)=ord(bab^-1).

1. Jun 16, 2012

### jmjlt88

Prove ord(a) = ord (bab-1).

Attempt at the proof.

Proof:

Let G be a group and let a,b ε G.

(=>)
Suppose ord(a)=n. That is, an=e. We wish to that the ord(bab-1)=n. If we take an=e and multiply on the left by b and on the right by b-1 then, we have,

(1) an=e
(2) banb-1=beb-1
(3) banb-1=e.
Hence, we have the fact that,
(4) an=e=banb-1.
Now, if we expand (bab-1)n, we get,
(5) bab-1bab-1bab-1..bab-1=banb-1.
But, banb-1 = e by our algebra in (1) - (4). Thus, (bab-1)n=banb-1 = e and ord(bab-1)=n.

(<=)
Let ord(bab-1)=n. That is, (bab-1)n=e. But, as it was shown in (5), (bab-1)n=banb-1 = e. Using the fact that banb-1 = e, we multiply on the left by b-1 and on the right by b and obtain that an=e. Hence, ord(a)=n.

QED

How does it look? Thanks!!

2. Jun 16, 2012

### Sajet

First of all, you can reduce everything you have done in "=>" to one line basically:

(bab-1)n=banb-1 = beb-1 (since ord a = n) = e.

But be aware that this has not yet shown ord(bab-1) = n. ord g is defined to be the smallest non-negative integer such that ...

Same goes for "<=".