Abstract algebra (If H is a subgroup of the additive group Z)

If H is a subgroup of the additive group Z containing (n)
where n∈N, then prove that H=(d) for some d∈N such
that d|n.
Note: N is positive integer

Note that Z is an infinite cyclic group.
So, H must be cyclic as well: H = (d) for some d∈N.

By hypothesis, H = (d) contains (n).
In particular, (n) is a subset of (d).
==> n is an element of (d)
==> n = kd for some k ∈ Z
==> d | n.

I hope this helps!

So Z is the integers, H is a subgroup. There must be some natural number d such that H is the set of multiples of d.

This may be obvious, but it isn't hard to prove. Let d be the smallest positive integer in H. Let h be any other number in H. Let m be the greatest common divisor of d and h. If m < h, there is are some integers a and b such that m = ad + bh, and therefore m is in H. But k was the smallest natural number in H, so this is a contradiction. Therefore m = d, and d is a divisor of k, for all k in H.

In particular, d is a divisor of the n given as being an element of H.

There are many ways to prove this, the choice depends on what you already know and can work with.