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## Main Question or Discussion Point

I'm looking to show that:

Order (ab(a^-1)) = Order b

So far...

Let x be order of ab(a^-1), so we have:

e = (ab(a^-1)^x)

= (a^x)(b^x)(a^(x-1))

= (a^x)(b^x)((a^-1)^x) so by associativity we have...

= (a^x)((a^-1)^x)(b^x)

= (((a)(a^-1))^x)(b^x)

= (e^x)(b^x)

= b^x

Hence x is the order of b aswell.

Im really not sure of how indices work with group elements, is it the same as with actual numbers? ie does (ab)^x = (a^x)(b^x) in groups? Im not sure it does!

Any help greatly appreciated!

Order (ab(a^-1)) = Order b

So far...

Let x be order of ab(a^-1), so we have:

e = (ab(a^-1)^x)

= (a^x)(b^x)(a^(x-1))

= (a^x)(b^x)((a^-1)^x) so by associativity we have...

= (a^x)((a^-1)^x)(b^x)

= (((a)(a^-1))^x)(b^x)

= (e^x)(b^x)

= b^x

Hence x is the order of b aswell.

Im really not sure of how indices work with group elements, is it the same as with actual numbers? ie does (ab)^x = (a^x)(b^x) in groups? Im not sure it does!

Any help greatly appreciated!