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All non-trivial proper subgroups of (R, +) are cyclic.

False

Discussion: There is a simple counter example: (Q, +) (the additive group of rational numbers). We also note that every additive subgroup of integers is cyclic (in fact they are of the for nZ).

Cyclic groups have exactly one generator. We can construct numerous counter examples by constructing subgroups having more than one generator. Hence another counter example is $a + b \sqrt 2$ which has two generators.

Why does this does not work for integers? For example can we say that {ax + by | a, b are given integers and x, y are arbitrary} form a subgroup with two generators? No because by Bezout’s Theorem we know that ax + by is merely the multiples (positive and negative) of the g.c.d(a,b).