Question: If $G$ is finite cyclic group then why is it true that for every divisor of the cardinality of $G$, there only exists at most one subgroup with such cardinality?
I am struggling to see a way to explain this properly. I realise that this group $G\cong \mathbb{Z}/n\mathbb{Z}$ and every subgroup of a cyclic group is again cyclic.
My attempt thus far:
Suppose for a contradiction that for some $d<n$ such that $\exists p,q\in G$ such that $ord(p)=ord(q)=d$ with the property that $\langle p \rangle \neq\langle q\rangle$ but now how do I proceed further? (I realise this eventually boils down to finding two elements with order $d$ such that one is not in the other's orbit.)
Moreover I saw that the converse is also true and thus could someone please give me some feedback for my attempt, too?
Claim I also tried to prove: Suppose for every $d<n=|G|$ there only exists one subgroup with cardinality $d$ then show $G$ is cyclic.
My attempt to this claim: Suppose $G$ is not cyclic then by structure theorem that $G\cong \mathbb{Z}/p_1\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}/p_n\mathbb{Z}$ where $p_1|\cdots|p_n$ and $n\geq 2.$ Now clearly, since $n\geq 2$ there are at least two subgroups with order $p_1$ and thus we reached a contradiction (for instance, since $\mathbb{Z}/p_1\mathbb{Z}$ is embedded in $\mathbb{Z}/p_2\mathbb{Z}$)
Many thanks in advance for providing any feedback to either question! 🙂
Best Answer
Hints: if $G=\langle g \rangle$, $|G|=o(g)=n$, and $d \mid n$, then $o(g^{\frac{n}{d}})=d$. Conversely, if $|\langle h \rangle|=d$, and $h=g^i$, then by Lagrange $d \mid n$. And $n|id$, whence $\frac{n}{d}|i$, say $i=m \cdot \frac{n}{d}$. So $h=(g^{\frac{n}{d}})^m$. Over to you to fill in the details. Note that this shows that for every divisor $d$ of $|G|$ there is precisely one subgroup of order $d$.
For your second question/claim: let $\varphi$ denote the Euler function, then if there is exactly one subgroup of order $d$ for every $d \mid n$, implies $G$ has at most $\varphi(d)$ elements of order $d$ for each $d\mid n$. But $\sum_{d\mid n}\varphi(d)=n$, so $G$ must contain exactly $\varphi(d)$ elements of order $d$ for each $d\mid n$, in particular, $G$ contains an element of order $n$.
Note Here is a different proof of the fact that if the finite group $G$ has the property
$(*)$ for each $d \mid |G|$ there is exactly one subgroup of order $d$
then $G$ is cyclic. Because of $(*)$ every Sylow subgroup is normal and hence $G$ is nilpotent, that is the direct product of its Sylow subgroups. If $|G|$ has more than one prime factor, by induction (and basically the Chinese Reaminder Theorem) we are done. So we can assume that $G$ is a $p$-group for some prime $p$. Pick a $g \in G$ of maximal order and assume $M=\langle g \rangle \lt G$. Let $x \in G-M$. Then, since $G$ is a $p$-group, we can find an element of $o(x)$ in $\langle g \rangle$. But owing to $(*)$, this forces $\langle x \rangle \subseteq M$, contradicting $x \notin M$. So $G=M$ is cyclic.