# 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 May 28 '15 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 $n$.

• I follow your suggestion. You are right. Thanks! – Hang May 31 '15 at 11:11
• Can you explain why do we have at most $\varphi(d)$ elements of order $d$? – Ninja Nov 22 '16 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 $d$. – Censi LI Nov 23 '16 at 17:35
• I still can not see the result, unfortunately. – Ninja Nov 23 '16 at 18:40
• So you meant $G$ has $φ(n)$ generators? – Nour Apr 29 '17 at 9:37

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.

Hint

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

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 $$n$$.

• your comment added even less other than make you feel superior – Guy Schwartzberg May 18 at 20:06

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 $$d$$.