No. The order of is another name for its cardinality. By the way, writing H=<g^d> would mean that is cyclic, which is what you are asked to prove.

For instance (there is a simpler proof that I can't remember), you can easily see that for any , so that, by induction, the subgroup spanned by any number of elements is always spanned by a single element, i.e. is cyclic. In particular, H is spanned by its elements, so that it is cyclic. ** this works if is finite only...

To prove the above identity, just write Bézout's identity where . Thus . And the reverse inclusion comes from and .

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A general proof: the function defined by is a group morphism. Thus, is a subgroup of , hence there is such that , hence is cyclic. This uses the knowledge of the subgroups of .

If is a nontrivial subgroup of , let be the smallest positive element of , then for every element , Euclidean division gives with , and we have hence by minimality of , and thus . qed.