- #1
sairalouise
- 10
- 0
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? I am 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? I am not sure it does!
Any help greatly appreciated!