Lemma $1.92$ in Rotman's textbook (Advanced Modern Algebra, second edition) states,
Let $G = \langle a \rangle$ be a cyclic group.
(i) Every subgroup $S$ of $G$ is cyclic.
(ii) If $|G|=n$, then $G$ has a unique subgroup of order $d$ for each divisor $d$ of $n$.
I understand how every subgroup must be cyclic and that there must be a subgroup for each divisor of $d$. But how is that subgroup unique? I'm having trouble understanding this intuitively. For example, if we look at the cyclic subgroup $\Bbb{7}$, we know that there are $6$ elements of order $7$. So we have six different cyclic subgroups of order $7$, right?
Thanks in advance.
Best Answer
Let $d$ be a divisor of $n=|G|$. Consider $H=\{ x \in G : x^d =1 \}$. Then $H$ is a subgroup of $G$ and $H$ contains all elements of $G$ that have order $d$ (among others).
If $K$ is a subgroup of $G$ of order $d$, then $K$ is cyclic, generated by an element of order $d$. Hence, $K\subseteq H$.
On the other hand, $x\in H$ iff $x=g^k$ with $0\le k < n$ and $g^{kd}=1$, where $g$ is a generator of $G$. Hence, $kd=nt$ and so $k=(n/d) t$. The restriction $0\le k<n$ implies $0\le t<d$, and so $H$ has exactly $d$ elements. Therefore, $K=H$.