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

[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!

 

[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 "<=".