3

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 $d\mid n,$ then $G$ is cyclic (and therefore has exactly one subgroup of each order $d\mid n$ ).

enter image description here

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?

Shaun
  • 44,997
Akira
  • 17,367
  • 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 '20 at 14:24

1 Answers1

2

I am confused by your question, so I apologize if my answer misses the point.

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 $n$.

Then it uses the divisor sum property of Euler's totient function.

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 $n<n$ which is a contradiction.

AnalysisStudent0414
  • 8,173
  • 1
    You are right. Thank you so much! – Akira Jun 27 '20 at 14:10
  • 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 '20 at 14:10
  • 1
    I see you have deleted your previous comment, but I'll leave the reply up in case someone else is confused in the future! – AnalysisStudent0414 Jun 27 '20 at 14:10