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

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.

• How much group theory do you know already? And is this a homework problem?
– KCd
Commented May 28, 2015 at 15:27

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 $n$.

– Hang
Commented May 31, 2015 at 11:11
• Can you explain why do we have at most $\varphi(d)$ elements of order $d$? Commented Nov 22, 2016 at 20:51
• @Ninja Each such element is a generator of a subgroup of order $d$, and by presumption $G$ has only one subgroup of order $d$. Commented Nov 23, 2016 at 17:35
• I still can not see the result, unfortunately. Commented Nov 23, 2016 at 18:40
• So you meant $G$ has $φ(n)$ generators?
– Nour
Commented Apr 29, 2017 at 9:37

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 $$d$$.

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.

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 $$n$$.