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If $G$ is a cyclic group of order $n$ and $k|n$, then $G$ has a subgroup of order $k$, it's easy to prove, but what I'm having troubles to show is there is only one subgroup of order $k$.

I need help here

Thanks a lot

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I bet this has been already asked here. – Mariano Suárez-Alvarez Apr 24 '13 at 8:00
Probably $>10$ times. – Martin Brandenburg Apr 24 '13 at 8:19

Consider the epimorphism $f\colon\mathbb Z\to G$ given by sending $1$ to a chosen generator of $G$. Then $\ker f=n\mathbb Z$. Show that for $H<G$ we have $f^{-1}(H)=m\mathbb Z$ (as that's how subgroups of $\mathbb Z$ look like) for some $0<m|n$ and that necessarily $m=[G:H]$.

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Hint: If $H$ is a subgroup of $G$ of order $k$, try to show that $H$ is the $k$-torsion of $G$, i.e. $H=\{g \in G : g^k=1\}$. Namely, $\subseteq$ follows from Lagrange, and $=$ follows when you show that the $k$-torsion has exactly $k$ elements. For that choose a generator of $G$ and write down the $k$-torsion explicitly.

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Let $G$ be cyclic of order $n$, let $k$ be a divisor of $n$ and let $K$ and $L$ be two subgroups of $G$ of order $k$. We want to show that $K=L$, and it is enough to show that the images of $K$ and $L$ are equal in $G/K\cap L$. It follows that we can suppose without loss of generality that $K$ and $L$ intersect trivially.Then the subgroup $K+L$ of $G$ is actually the direct sum of $K$ and $L$, and since these two are cyclic of the same order, $K+L$ is not cyclic —as it must be, because all subgroups of $G$ are cyclic­— unless $k=1$, and then of course $K=L$.

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We prove this by induction on $n=|G|$. Of course this holds when $n=1$, so assume $n>1$ and for all $m<n$ the cyclic groups of order $m$ have this property.

Suppose $H$ and $K$ are subgroups of $G$ of order $k$. Then $HK$ is a cyclic subgroup of $G$. If $|HK|<n$, then the inductive hypothsis implies $H=K$. On the other hand, assume $HK=G$. Let $g$ be a generator of $G$ and write $H=\langle g^a\rangle$, $K=\langle g^b\rangle$. Because $|H|=k$, $n\mid ak$ so that $n/k$ divides $a$. Similarly, $n/k$ divides $b$. However, unless $n/k=1$, this absurd since $g$ must be expressible in the form $g^{ra+sb}$. It follows that $H=G=K$.

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Why must both sugroups contain generators of $G$ in the last case? – Mariano Suárez-Alvarez Apr 24 '13 at 8:37

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