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 $d$ | $n$ ).

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?

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

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ You are right. Thank you so much! $\endgroup$ – LAD Jun 27 at 14:10
  • $\begingroup$ 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. $\endgroup$ – AnalysisStudent0414 Jun 27 at 14:10
  • $\begingroup$ I see you have deleted your previous comment, but I'll leave the reply up in case someone else is confused in the future! $\endgroup$ – AnalysisStudent0414 Jun 27 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.