Long abstract algebra problem?
Let G be a finite abelian group. Suppose a,b are in G, a is not equal to b, and no proper subgroup of G contains {a,b}.
If G is not equal to <a>, and G is not equal to <b>, then G is isomorphic
More Questions From This User See All
Answers & Comments
Verified answer
If G isn't <a> then b is not in <a>, and if G isn't <b> then a isn't in <b>. In other words <a>∩<b> = 1. Since no proper subgroup of G contains {a,b}, the group generated by a and b, and since G is abelian then G = {a,b} ≈<a>x<b>≈Z_m x Z_n, where m = ord(a) and n = ord(b).