Question about Cyclic Group 
$(G,\star)$ 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.
 A: 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 $d=1$.
A: 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 $n$.
