A finite group which has a unique subgroup of order $d$ for each $d\mid n$.

Question

Problem Suppose G is a finite group of order $n$ which has a unique subgroup of order $d$ for each $d\mid n$. Prove that $G$ must be a cyclic group.

My idea: I try to prove it by induction. Let $p|n$ be a prime. Then by condition, there exists a unique subgroup $H$ of order $n/p$. Since $|gHg^{-1}|=|H|$, we must have $gHg^{-1}=H$ by the uniqueness part of the condition. So, H is a normal subgroup. Now, $|G/H|=p$ and thus $G/H=\langle x\rangle$ where $x^p \in H$.

However, I cannot continue.

My another idea is that first consider the case when $|G|$ is a power of some prime $p$. But, it still doesn't work.

Answer 1

Let $\varphi$ denote the Euler function, then you constraint forces that $G$ 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 $n$.

Answer 2

Here is an example, which shows in particular how Censi LI's answer works out. Suppose $G$ is a group of order $12$, which has only ONE subgroup of orders $1,2,3,4$ and $6$. Well, the only subgroup of order $1$ is, of course, $\{e\}$, so we have $11$ elements left. We have a single subgroup of order $2$, which is of the form $\{e,a\}$ for some element $a$ of order $2$. Now we have $10$ elements left.

Since we have a subgroup of order $3$, we have two more elements of order $3$ (indeed, our subgroup must be $\{e,c,c^{-1}\}$ for some element $c$ of order $3$). Now we have $8$ elements left. The subgroup of order $4$ is a bit more interesting:

Firstly, it must be cyclic, for a non-cyclic subgroup of order $4$ would have $3$ elements of order $2$, giving rise to $3$ subgroups of $G$ of order $2$, and $G$ only has one such subgroup. So our subgroup must be $\{e,d,a,d^{-1}\}$, for some element $d$ of order $4$ (with $d^2 = a$). This has $2$ elements of order $4$ ($d$ and $d^{-1} = d^3$), leaving $6$ elements left to account for.

Next, we have a subgroup of order $6$: it might be (hypothetically) that this subgroup is non-abelian, but then it would be isomorphic to $S_3$ which has $3$ (which is too many) elements of order $2$. So it must be an abelian group of order $6$, which is cyclic, and is actually: $\{e,f,c,a,c^{-1},f^{-1}\}$, where:

$f^2 = c,f^3 = a,f^4 = c^{-1} = c^2,f^5 = f^{-1}$

for some element $f$ of order $6$. We see that $f^{-1} = f^5$ is also of order $6$, which thus accounts for two "new" elements we haven't encountered before. This leaves $4$ elements left-over, which must be of order $12$, since all the lower orders are already accounted for.

As you can see, these subgroups each have $\phi(d)$ elements of order $d$, for $d = 1,2,3,4,6$. And lo and behold:

$1 + 1 + 2 + 2 + 2 + 4 = 12$, so this is all elements.

In other words, the Euler totient function, $\phi$, acts as a kind of "seive" weeding out the lower orders as we move through the divisors of the order of our group.

Answer 3

Keith Conrad (citing Trevor Hyde for part of the proof) gives a proof in Appendix A of this article. Here is an outline of the proof:

Proposition. Let $G$ be a finite group. Suppose that for each positive divisor $d$ of $|G|$, there is at most one subgroup of order $d$. Then $G$ is cyclic.

Proof. First suppose that $|G|=p^m$, where $p$ is a prime. Let $g\in G$ be an element with a maximum order. Let $h\in G$ be any element. Let $|g|=p^k$ and $|h|=p^\ell$. Then $p^k\geq p^\ell$, so $p^\ell\mid p^k$. Since $\langle g\rangle$ is cyclic, it has a (unique) subgroup of order $p^\ell$. But $\langle h\rangle$ has order $p^\ell$ as well, so our assumption implies that they are the same set. Hence, $\langle h\rangle\subseteq\langle g\rangle$, and in particular $h\in\langle g\rangle$. Since $h$ was arbitrary, we then have $G\subseteq\langle g\rangle$, i.e., $G=\langle g\rangle$.

Next, suppose that $|G|=p_1^{m_1}\cdots p_r^{m_r}$, where $p_1,\ldots,p_r$ are distinct primes. Let $H_i\in\operatorname{Syl}_{p_i}(G)$. Our assumption implies that $n_{p_i}=1$ for each $i=1,\ldots,r$, so $$G\simeq H_1\times\cdots\times H_r.$$ On the other hand, by the first part of the proof, $H_i\simeq\mathbb{Z}_{p_i^{m_i}}$ for each $i=1,\ldots,r$. By the Chinese remainder theorem, $G$ is cyclic.

Corollary. Let $G$ be a finite group. TFAE:

1. $G$ is cyclic.
2. For each positive divisor $d$ of $|G|$, there is at most one subgroup of order $d$.
3. For each positive divisor $d$ of $|G|$, there exists a unique subgroup of order $d$.

Answer 4

Let $G = \{a_1, a_2, \dots, a_n\}$, define $H = (a_1, a_2, \dots, a_n)$. Then taking any element in $H$, we compute $|\langle a_i \rangle| (= k \text{ say})$, then, since we have unique subgroup of order $k$, we have $\phi(k)$ such elements in total in $G$, we will remove such elements from $H$ and continue like this until $H$ becomes empty. Now since order of a subgroup always divides order of the group and the fact that $\sum_{d | n}\phi(d) = n$, we must exaust all the divisors of $n$ and thus we will have an element of order $n$.

Answer 5

Hint

Count the elements of each possible order. Conclude that there must be an element of order |G|.