Question

*GROUP THEORY/ABSTRACT ALGEBRA*

If a ∈ G and a^m = e, prove that o(a) | m

Answer 1

Order of an element: If G is a group and a∈G is an element of G,then a is of order n if n is the least positive integer such that aⁿ = e................................................(1)

Given a^m = e ........................................(2)

If possible let o(a) i.e. n does not divide m

then using division algorithm we can write

m = nq + r where q and r are integers and 0 < r < n

[ Note: we have only left the possibility of r = 0, which is true when n | m ]

or a^m = a^{nq + r}

or e = (aⁿ)^q (a)^r { using (2)}

or e = e^q (a)^r { using (1)}

or a^r = e

but r < n which means that there exists a positive integer less than n such that a^r = e. This contradicts (1) as n was the least postive integer.

Therefore our assumption was wrong and r = 0

or n | m

or o(a) | m