$(G,\star)$ is a $n$-order cyclic group, the generator is $x$, Prove:
1) For any factor $d$ of $n$ exists only one $d$-order subgroup.
2) $b=x^k$ is $G$'s generator if and only if $\gcd(n,k)=1$.

1)Suppose $n=d\cdot m$, then $a^m$ 's order is $d$, but I don't know how to prove there is only one subgroup.
2)I feel the question is too hard for me, but really want to solve it. :)

Any ideas? thanks.

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Hints: 1) Let $H$ be a subgroup of order $d$. All elements of $H$ are of the form $x^k$ for some $0\le k< |G|$. Think of the smallest $k$ 2) what is the order of $b$? –  yohBS Apr 8 '12 at 17:16
I am reasonably sure this is a duplicate--I'll try if I can find one. –  user21436 Apr 8 '12 at 17:18

For (1), suppose that $H$ is a subgroup of $G$ of order $d$. Let $y\in H$. Since $G$ is cyclic with generator $x$, you know that $y=x^k$ for some integer $k$ such that $0\le k<n$. Show that $k$ is a multiple of $m$ and conclude that $y$ is in the subgroup generated by $x^m$. Since $y$ was an arbitrary element of $H$, it follows that $H$ is contained in the subgroup generated by $x^m$, and since these sets both have $d$ members, they must be equal. There is a standard trick to proving that $k$ is a multiple of $m$: use the division algorithm to write $k=mq+r$, where $q$ and $r$ are integers and $0\le r<m$, and show that $r$ must be $0$. To do that you’ll want to use the fact that since $H$ has order $d$, $y^d=1_G$.

For (2), you know that $b$ is a generator of $G$ if and only if the order of $b$ is $n$. You know that $y^n=1_G$ for every $y\in G$, so the order of $b$ is $n$ if and only if no smaller positive power of $b$ is the identity. Thus, you want to show two things:

$\qquad\qquad\qquad(a)\quad$ if $\gcd(n,k)=1$, and $0<m<n$, then $b^m\ne 1_G$,

and

$\qquad\qquad\qquad(b)\quad$ if $\gcd(n,k)>1$, then the order of $b$ is less than $n$.

The first implies that whenever $\gcd(n,k)=1$, $b$ is a generator of $G$; the second implies that whenever $\gcd(n,k)>1$, $b$ is not a generator of $G$.

For $(b)$, think about what you did in problem (1). For $(a)$, you need to show that $km$ is not a multiple of $n$.

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1) Define a homomorphism $\varphi$: $\mathbb{Z}\rightarrow G$ such that $\varphi(n)=x^m$. Then $\varphi$ is surjective. Consider the kernel of $\varphi$, $N=\ker\varphi$. Since $G$ is of order $n$, $N=n\mathbb{Z}.$ By the isomorphism theorem, $$G\cong\mathbb{Z}/n\mathbb{Z}.$$ Applying the isomorphism theorem again, the subgroups of $G$ corresponds to the subgroups $t\mathbb{Z}$ in $\mathbb{Z}$ such that $t|n,~s>0.$ The image of these subgroups in $G$ are $\langle x^t\rangle$ with order $n/t= d$.

2) I will show a more general case. $\langle x^k\rangle$ is a subgroup of $G$. As 1) shows, $\langle x\rangle$ has subgroups of the form $\langle x^d\rangle$ with $d|n$. Since $\langle x^k\rangle=\langle x^d\rangle$, we have $$k=ad+bn,$$ $$d=uk+vn,$$ where $a,b,u,v$ are integers. Then we obtain that $d|k$ and $d=(n,k)$. Applying 1), we have the order of $\langle x^k\rangle$ is $n/d$, so is the order of $x^d$. Your question is the case for $d=1$.

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For question $1$, you can start with the following facts:

$(1)$ Every subgroup of a cyclic group is cyclic.

$(2)$ If $g$ is an element of order $n$ and $d$ divides $n$, then $g^d$ has order $n/d$.

$(3)$ If $g$ has order $n$, then $\langle g^k \rangle = \langle g^{\gcd(n, k)} \rangle$ for any $k \in \mathbb{Z}$. (To prove this, use Bezout)

Now let $\langle g \rangle$ be a cyclic group of order $n$ and $d$ some divisor of $n$. Then $\langle g^{n/d} \rangle$ is a subgroup of order $d$ by $(2)$. To show that it is the only subgroup of order $d$, let $H$ be some subgroup of order $d$. Then $H = \langle g^k \rangle$ for some $k$ dividing $n$ by $(1)$ and $(3)$. Because $g^k$ has order $d$, we have $n/k = d$ by $(2)$. Therefore $H = \langle g^{n/d} \rangle$, and the proof is complete.

For question $2$, let $\langle g \rangle$ be a cyclic group of order $n$. Then if $g^k$ is a generator, we have $g = g^{kq}$ for some integer $q$. Thus $g^{kq - 1} = 1$ and $n$ divides $kq - 1$. Therefore $kq - np = 1$ which implies $\gcd(k,n) = 1$.

For the other direction, we use Bezout. If $\gcd(k,n) = 1$, then $xk + yn = 1$ for integers $x$ and $y$. Thus $g^i = g^{xni + yki} = (g^{k})^{yi}$ since $g^n = 1$.

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