1. ## theorem

im trying to prove that the subgroup of a cyclic group is always cyclic but im wondering what is wrong with my proof.

what i did:

let G=<g>={g^n l n is an integer} and let H be the subgroup with order d. then d is the least positive integer such that g^d=e.
thus d is a divisor of n.
since the order of g^n =G is the order of H =d then d=n and hence H=<g>

im wondering what is wrong with my understanding because in my notes, it states that to proof this theorem, i need to let H=<g^d> ={(g^d)^q} and im wondering why i need to do it that way.

buy letting H =<g^d>, it still means that it has an order d right?

2. Originally Posted by alexandrabel90
buy letting H =<g^d>, it still means that it has an order d right?
No. The order of $H$ is another name for its cardinality. By the way, writing H=<g^d> would mean that $H$ 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 $\langle g^m,g^n\rangle = \langle g^{\gcd(m,n)}\rangle$ for any $m,n\in\mathbb{Z}$, 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 $G$ is finite only...

To prove the above identity, just write Bézout's identity $am+bn=d$ where $d=\gcd(m,n)$. Thus $g^d=(g^m)^a(g^n)^b\in \langle g^m,g^n\rangle$. And the reverse inclusion comes from $d|m$ and $d|n$.

--
A general proof: the function $\phi:\mathbb{Z}\to G$ defined by $\phi(n)=g^n$ is a group morphism. Thus, $\phi^{-1}(H)$ is a subgroup of $\mathbb{Z}$, hence there is $a\in\mathbb{Z}$ such that $\phi^{-1}(H)=a\mathbb{Z}$, hence $H=\phi(a\mathbb{Z})=\langle g^a\rangle$ is cyclic. This uses the knowledge of the subgroups of $\mathbb{Z}$.
If $H$ is a nontrivial subgroup of $\mathbb{Z}$, let $h$ be the smallest positive element of $H$, then for every element $a\in H$, Euclidean division gives $a=qh+r$ with $0\leq r, and we have $r=a-qh\in H$ hence $r=0$ by minimality of $h$, and thus $a=qh\in h\mathbb{Z}$. qed.

3. then in this case, how do i find out the order of H=<g^d>?