# If a finite group $G$ of order $n$ has at most one subgroup of each order $d|n$, then $G$ is cyclic

I'm reading the proof of a theorem in Fundamentals of Group Theory An Advanced Approach by Steven Roman.

(Characterization by subgroups) If a finite group $$G$$ of order $$n$$ has the property that it has at most one subgroup of each order $$d$$ | $$n,$$ then $$G$$ is cyclic (and therefore has exactly one subgroup of each order $$d$$ | $$n$$ ).

Here $$o(a)$$ is the order of $$a$$ and $$\phi(d)$$ is the number of elements of order $$d$$.

Because the order of a subgroup must be a divisor of that of a group, I get $$n=\sum_{d \in D} \phi(d) = \sum_{d \mid n} \phi(d)$$ Then I'm stuck at getting how $$\phi(n) > 0$$.

Could you please elaborate on this point?

• You could view the right hand side of the inequality as applying the left hand side to the cyclic group $\mathbb Z/n$. – Justin Young Jun 27 at 14:24

The proof you quote is using Lagrange's theorem in the "$$\leq$$" step, that is, that the order of an element ($$d$$) has to divide the order of the group ($$n$$), and so without loss of generality you can restrict the sum only to those $$d$$ that divide $$n$$.
En passant, this also proves that there is exactly one subgroup of order $$d$$ (opposed to the "at most one" of the hypothesis), because otherwise you'd get $$n which is a contradiction.
• Not really. The author takes an element $a$ of order $d$, and argues that $\langle a \rangle$ is a subgroup of order $d$, and it is hence unique (by the hypothesis). Then all elements of order $d$ are contained in that subgroup. Now, this subgroup is cyclic (generated by $a$) and a cyclic group of order $d$ contains exactly $\phi(d)$ elements of order exactly $d$, where $\phi$ is Euler's totient function. – AnalysisStudent0414 Jun 27 at 14:10