# If an Abelian group $G$ has order $n$ and at most one subgroup of order $d$ for all $d$ dividing $n$ then $G$ is cyclic

If an Abelian group $G$ has order $n$ and at most one subgroup of order $d$ for all $d$ dividing $n$ then $G$ is cyclic.

I am trying to use the structure theorem for finitely generated abelian groups.

So I write $n=p_1^{\alpha_1}\ldots p_n^{\alpha_n}$.

I am hoping to show each of the alpha's must =1 then I will have that $G$ is isomorphic to $\prod_i^n \mathbb{Z}/p_i \mathbb{Z}$, which is cyclic.

Just the most basic group theory and a simple counting argument will suffice. By Lagrange's theorem $G$ only has elements of orders $d$ dividing $n$. If such an element exists, then the cyclic group it generates has a number $\phi(d)$ of elements of order $d$ that only depends on $d$ (this is Euler's totient function but we need to know nothing about it). By the hypothesis $G$ cannot contain other elements of order $d$, so it has either $\phi(d)$ such elements or none; set $\chi_G(d)=1$ if it does have such elements and $\chi_G(d)=0$ if it doesn't. Then counting element of $G$ by their order one has $$n=\#G=\sum_{d\mid n}\chi_G(d)\phi(d).$$

But a cyclic group of order $n$ has exactly one cyclic subgroup of order $d$ for every $d$ that divides $n$. This means that $$\sum_{d\mid n}\phi(d)=n.$$ Thus the formula for $G$ can only be satisfied if all values $\chi_G(d)$ are equal to $1$. In particular $\chi_G(n)=1$: the group $G$ contains elements of its own order $n$ and is therefore cyclic.

• @MarcvanLeeuwenAs a beginning self-studier, maybe I could ask you a few questions since this answer is pertinent to a question I asked regarding G=25. How can I prove the statement in the middle "But a cyclic group..." in both directions? And I would appreciate help with the last summation since if d = n, you get n elements; so what happens to the sum as you add the additional d elements for each d that divides n? Thanks this would be a big help. – user12802 Jan 7 '12 at 21:25
• @MarcvanLeeuwenMaybe you might want to post it as an answer to my question math.stackexchange.com/questions/96985/… – user12802 Jan 7 '12 at 21:33
• @Andrew: In the last summation $d=n$ accounts for only $\phi(n)$ elements, which is less than $n$. For the middle statement, if $C$ is cyclic of order $n$ with a generator $g\in C$, and $d$ is a divisor of $n$, then $g^{n/d}$ generates a cyclic subgroup of order $d$. Any cyclic subgroup $C'$ of $C$ has this form, since if $i>0$ is minimal with $g^i\in C'$ then $i$ divides $n$: if it didn't the remainder of $n$ divided by $i$ would be a number contradicting the minimality of $i$. The $\phi(n)$ different generators of $C$ are what is left after removing the union of all strict cyclic subgroups. – Marc van Leeuwen Jan 7 '12 at 22:11
• @MarcvanLeeuwenThanks,especially to respond so close to midnight in France. – user12802 Jan 7 '12 at 22:34

Let $\phi$ be Euler's totient function.

The assumption implies that $G$ has at most $\phi(d)$ elements of order $d$ whenever $d$ divides $n$.

But then it has at least $\phi(d)$ elements of order $d$ whenever $d$ divides $n$.

So it has elements of order $n$.

EDIT. As Thomas suggests, I should add that the following fact has been implicitly used: We have $$\sum_{d|n}\ \phi(d) = n$$ by considering the case when $G$ is cyclic.

• You should note: By the theorem: $\sum_{d|n}\phi(d) = n$ – Thomas Andrews Dec 22 '11 at 17:59
• Dear @ThomasAndrews: Thank you very much! – Pierre-Yves Gaillard Dec 22 '11 at 18:04

I would prove this along the lines of J Rotman - An introduction to the theory of Groups, Graduate Texts in Mathematics, Springer-Verlag, 4th Edition. Pg:28

Observe that the converse of what you'd like to prove is quite a trivial issue.

For what you'd like to prove, J R has the following Recipe:

Lemma: Let $\phi$ be the Euler's totient function that counts the number of numbers less than $n$ and coprime to it, $$\sum_{d|n} \phi (d) = n$$

Proof:

The proof of this fact follows from the following: Let G be a cyclic group of order $n$. I'll assume you can prove that there are $\phi(n)$ generators for this $G$ and in general $\phi(k)$ generators for a cyclic group of order $k$.

$$G=\bigcup gen(C)$$ where $gen(C)$ stands for generators of cyclic subgroups $C$ of G.(So $C$ ranges over all cyclic subgroups of $G$) And this union is a disjoint union, from the definition of a generator.

Note that the disjoint union holds for all $G$ and there is nothing sacrosanct about $G$ being cyclic, except from here: I claim now that, as the subgroups are unique for each divisor, the disjoint union translates into a sum that is exactly the lemma we seek to prove.

As pointed out before, I'll make use of the disjoint union.

$$n=|G| = \sum |gen(C)| \leq \sum_{d|n}{\phi(d)}$$

So, now I use my lemma to conclude that, I have cyclic subgroups for each divisor of $n$ and in particular for $n$. So, that must be $G$. So, G must be cyclic.

I would like to emphasise, that this classifies all finite cyclic groups.

• Will the downvoter care to explain why he downvoted this answer? – user21436 Jan 23 '12 at 19:31

It may be easier if you use the form of the structure theorem that says that $G$ is a product of cyclic groups where the order of each divides the order of the next.

• I think it must be a comment not a answer!! – user21436 Dec 22 '11 at 17:36
• @Kannappan, you are entitled to your opinion. Personally, I think it is neither a comment nor an answer, but a hint - but there is no such category on this website, so I flipped a coin and went for answer. – Gerry Myerson Dec 22 '11 at 17:54
• @Kannappan: It would be more polite if you said "I think it ought to be a comment". Using the word "must" implies an urgency and insistence which is not warranted here. – Zev Chonoles Dec 22 '11 at 19:19
• @Zev Sorry about the language, not intentional at all. All I wanted to convey was in much lighter sense the fact that, this post doesn't answer the question – user21436 Dec 22 '11 at 19:21

By Cayley Theorem $G$ has an element $x$ of order $p_i$. It follows from your requirement that the subgroup $\langle x\rangle$ generated by $\langle x\rangle$ contains all the elements of order $p_i$.

Exercise for you: Prove that $p_i$ cannot divide the order of the factor group $G/\langle x\rangle$ (Why?)

and you are done...

P.S. If you covered this topic already, looking to the $p$-Sylow subgroups also helps, but I think this is more complicated. :)

• Wouldn't that imply that $n$ is squarefree, although any cyclic group satisfies the requirement? – Paolo Capriotti Dec 22 '11 at 18:50