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I am looking at this proof and I am stuck on the logic that $a^{p}$ = 1. For example, consider the group under multiplication without zero, ${Z}_{5}$, wouldn't 2^4 = 1 imply that the order is 4 not 5? We know that if G is a finite abelian group, G is isomorphic to a direct product ℤ(p1)^n1×ℤ(p2)^n2×⋯×ℤ(pr)^nr where pi's are prime not necessarily distinct. Consider each of the ℤ(pi)ni as a cyclic group of oreder pnii in multiplicative notation. Let m be the lcm of all the pnii for i=1,2,…,r. Clearly m≤p1n1p2n2⋯prnr. If ai∈ℤ(pi)ni then (ai)(pini)=1 and hence ami=1. Therefore for all α∈G, whe have αm=1 that is every element of G is a root of xm=1. However G has p1n1p2n2⋯prnr elements, while the polynomial xm−1 can have at most m roots in F. So, we deduce that m=p1n1p2n2⋯prnr. Therefore pi's are distinct primes, and the group G is isomorphic to the cyclic group ℤm.