# Group theory finding order of element and inverse

## Homework Statement

Let a and b be elements of a group,with a^2=e , b^6=e and a.b=b^4.a find its order and express its inverse in form of a^m.b^n

## The Attempt at a Solution

(ab)^2=(ab)(ab)=(ab)(b^4.a)=a(b^5)a
(ab)^3=a(b^5)a(ab)=a(b^5)(a^2)b=a(b^6)=ae=a
it implies(ab)^6=e
thus order(ab) divides 6.It is not 3 hence it is 6.

Help me in solving second part.I can express inverse in form of b^m.a^n but not in the asked form.

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you want (ab)^(-1)?

try (b^4a)^(-1)

## Homework Statement

Let a and b be elements of a group,with a^2=e , b^6=e and a.b=b^4.a find its order and express its inverse in form of a^m.b^n

## The Attempt at a Solution

(ab)^2=(ab)(ab)=(ab)(b^4.a)=a(b^5)a
(ab)^3=a(b^5)a(ab)=a(b^5)(a^2)b=a(b^6)=ae=a
it implies(ab)^6=e
thus order(ab) divides 6.It is not 3 hence it is 6.

Help me in solving second part.I can express inverse in form of b^m.a^n but not in the asked form.
(ab)^-1=(b^4 a)^-1 as algebrat has said.

Since you can find (ab)^-1 in the form b^m a^n, you should have no problem expressing (b^4 a)^-1 in the form a^m b^n.

you want (ab)^(-1)?

try (b^4a)^(-1)
thanks i have got it
if my solution to finding the order of ab correct or not

So you deduced that (ab)^2 = a b^5 a
If you raise that to the third power, you get a b^15 a.

So apparently e= (ab)^6 = a b^15 a.
Also, b^4=e

Hmmm.......

a=sqrt(e)

b=6th root of e => a=b^3

=>(ab)=b^4=>(b^4)^(-1)=b^(-1/4)=e^(-1/24)