# 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)

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This will only hold if $C$ is a finite cyclic group. E.g., $|2\mathbb Z| = |4\mathbb Z|$ but these subgroups of $\mathbb Z$ groups are not equal. – amWhy May 1 '13 at 18:42
@amWhy, of course if one reformulates the question in terms of indices, then it works for the infinite cyclic group as well. – Andreas Caranti May 1 '13 at 18:46
I agree, @AndreasCaranti – amWhy May 1 '13 at 18:47

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.

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I think the question would be boringly trivial if the OP knew/could use this theorem... – DonAntonio May 1 '13 at 19:08
i agree :) @donantonio – wanderer May 1 '13 at 21:45

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.

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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 $G$: $$\langle e_G\rangle,\langle a\rangle,\langle a^{2}\rangle,...,\langle a^{k}\rangle,...$$

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+1 Nice to have stated explicitly the theorem we're looking at! – amWhy May 3 '13 at 0:08