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Thread: theorem

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by alexandrabel90 View Post
    buy letting H =<g^d>, it still means that it has an order d right?
    No. The order of $\displaystyle H$ is another name for its cardinality. By the way, writing H=<g^d> would mean that $\displaystyle H$ 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 $\displaystyle \langle g^m,g^n\rangle = \langle g^{\gcd(m,n)}\rangle$ for any $\displaystyle m,n\in\mathbb{Z}$, 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 $\displaystyle G$ is finite only...

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

    --
    A general proof: the function $\displaystyle \phi:\mathbb{Z}\to G$ defined by $\displaystyle \phi(n)=g^n$ is a group morphism. Thus, $\displaystyle \phi^{-1}(H)$ is a subgroup of $\displaystyle \mathbb{Z}$, hence there is $\displaystyle a\in\mathbb{Z}$ such that $\displaystyle \phi^{-1}(H)=a\mathbb{Z}$, hence $\displaystyle H=\phi(a\mathbb{Z})=\langle g^a\rangle$ is cyclic. This uses the knowledge of the subgroups of $\displaystyle \mathbb{Z}$.
    If $\displaystyle H$ is a nontrivial subgroup of $\displaystyle \mathbb{Z}$, let $\displaystyle h$ be the smallest positive element of $\displaystyle H$, then for every element $\displaystyle a\in H$, Euclidean division gives $\displaystyle a=qh+r$ with $\displaystyle 0\leq r<h$, and we have $\displaystyle r=a-qh\in H$ hence $\displaystyle r=0$ by minimality of $\displaystyle h$, and thus $\displaystyle a=qh\in h\mathbb{Z}$. qed.
    Last edited by Laurent; Mar 14th 2010 at 12:04 PM.
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  3. #3
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    then in this case, how do i find out the order of H=<g^d>?
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