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I have this theorem in my abstract algebra book that the book doesn't prove so I was wondering what is the best way to prove this theorem I proved it by the fact that if we consider c an element of G of largest order then we must have a element of G that divides it, but however I don't like this proof since it relies on some polynomial being solvable.

  • You mean finite subgroup of anyone....? – Mikasa Mar 20 '15 at 16:05
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    Typically the proof uses the fact that the polynomial $x^n-1$ has atmost $n$ roots over a field. This is not the same as being "solvable". – Prahlad Vaidyanathan Mar 20 '15 at 16:09
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    Since $\mathbb{Q} \leq \mathbb{R} \leq \mathbb{C}$, I would focus solely on the complex numbers (I hope the $\mathbb{Z}_p$ part is obvious). It might help that, being an abelian group, every subgroup is the kernel of some homomorphism, but I'm not certain it will. To have a finite kernel, the image must be infinite in size, but that's not sufficient. – pjs36 Mar 20 '15 at 16:12

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Actually, the assertion is true for any field. It is based on the following result:

Lemma. If a finite group $G$ of order $n$ has at most $1$ subgroup of order $d$ for each divisor $d$ of $n$, then $G$ is cyclic.

Indeed, all elements of order $d$ of a finite subgroup of a field $K$ satisfy the equation $x^d-1=0$. As we're in a field, this equation has at most $d$ roots. Suppose there is an element $a$ of order $d$ in $K$. The equation $x^d = 1$ has $d$ roots, the elements of the subgroup generated by $a$: $\,\langle a\rangle=\{1,a, \dots, a^{d-1}\}$.

If there were another subgroup of order $d$, it would contain an element $b\notin\langle a\rangle$; its order would be a divisor of $d$ by Lagrange's theorem, so that $b$ would be a $d+1$-th root of the equation $x^d=1$.

Proof of the lemma:

Let $\varphi$ be Euler's totient function. If $a$ is an element of order $d$ in $G$, there are $\varphi(d)$ elements of order $d$ in $\langle a\rangle$. Let $r(d$ be the number of cyclic subgroups of order $d$ ($r(d)=0$ or $1$ by hypothesis). The total number of elements order $d$ in $G$ is $r(d)\varphi(d)$, hence: $$n=\lvert G\rvert=\sum_{d\mid n}r(d)\varphi(d)\le\sum_{d\mid n}\varphi(d)=n $$ so that $r(d)=1$ for all $d$; in particular $r(n)=1$, which proves $G$ is cyclic.

Bernard
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