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!