- #1
BomboshMan
- 19
- 0
Hi,
I can't understand why the statement in the title is true. This is what I know so far that is relevant:
- A subgroup of a cyclic group G = <g> is cyclic and is <g^k> for some nonnegative integer k. If G is finite (say |G|=n) then k can be chosen so that k divides n, and so order of g^k is n/k.
- Let G = <g> be a cyclic group. If g has infinite order, then G is isomorphic to the integers with addition. If o(g) = n, then G is isomorphic to Z/nZ with addition.
And then apparently the last point implies the statement in my title and I have no idea how!
Thanks,
Matt
I can't understand why the statement in the title is true. This is what I know so far that is relevant:
- A subgroup of a cyclic group G = <g> is cyclic and is <g^k> for some nonnegative integer k. If G is finite (say |G|=n) then k can be chosen so that k divides n, and so order of g^k is n/k.
- Let G = <g> be a cyclic group. If g has infinite order, then G is isomorphic to the integers with addition. If o(g) = n, then G is isomorphic to Z/nZ with addition.
And then apparently the last point implies the statement in my title and I have no idea how!
Thanks,
Matt