Subgroups of cyclic groups with same order Let $C$ be a cyclic group. Let $A$ and $B$ be two subgroups of $C$ with $|A|=|B|$. Then $A = B$.
How to show this? Thanks.
(Btw, I already know $A$ and $B$ are cyclic)
 A: 
Theorem 1: If $G=\langle a\rangle$ be a finite group of order $n$ and $$d_1,d_2,...,d_k$$ be all distinct positive divisors of $n$ so the following subgroups are all the proper distinct subgroups of $G$: $$\langle a^{d_1}\rangle,\langle a^{d_2}\rangle,...,\langle a^{d_k}\rangle $$
Theorem 2: If $G=\langle a\rangle$ be an infinite group  then the following subgroups are all the proper distinct subgroups of $G$: $$\langle e_G\rangle,\langle a\rangle,\langle a^{2}\rangle,...,\langle a^{k}\rangle,... $$

A: Hint: As amWhy noted you need $C$ to be finite, so you can assume $C = \mathbb Z/n$ for some $n$.  Let $a$ be the least positive integer contained in $A$.  Show that $a \mid n$ and consequently $a$ generates $A$.  Then $a = n/|A|$ is uniquely identified by the size of $A$, therefore $A$ is uniquely identified by it's size.
A: Recall that for every divisor of the order of a cyclic group $\exists$ a unique subgroup of that order. Now if $A$ and $B$ are two subgroups of a finite cyclic group $G$ then their orders must be divisors of the order of the group and due to uniqueness they must be the same.
