For a finite group G, I need to prove that the order of an element in G is a divisor of the order of the group. I'm not sure what this exaclty means, but I think you have to use cyclic groups such that G={a^0,a^1,.....,a^n) where n+1 is the order of the group and a^0 is the identity element. So I think I need to use Lagrage's thm stating that the order of a subgroup is a divisor of the order of the group. So to find a subgroup of G seems to be a problem. Could I use a^m as a subgroup, where m < n+1?? What is exaclty meant by order of the element. Thanks.(adsbygoogle = window.adsbygoogle || []).push({});

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# Groups Theory Proof

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