Let $G$ be a group, $|G|=n$. Suppose $\forall d\in \mathbb{N}$ such that $d\mid n$, there are at most $d$ elements s.t. $x^d=1$. Prove $G$ is cyclic.

Question

Let $G$ be a group, $|G|=n$. Suppose $\forall d\in \mathbb{N}$ such that $d\mid n$, there are at most $d$ elements such that $x^d=1$. Prove $G$ is cyclic.

I saw these posts:

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

• How do I show that a finite group $G$ of order $n$ is cyclic if there is at most one subgroup of order $d$ for each $d\mid n$?

I am not sure these questions' posts have the same meaning as my own question.

My attempt:

My approach is to show $G$ has at most one subgroup of each of each order $d\mid n$.

Using cauchy theorem there is a cyclic subgroup in order $p$ for each $p\mid n$ such that $p$ is a prime number.

Let $H<G,|H|=p$, $\forall h\in H$ exists $\space O(h)=p$ (Lagrange's Theorem).

Combining this and the given fact that there are at most $p$ elements $\implies H$ is the unique subgroup with order $p$.

My problem is, how do I deal with non-primary order subgroup?

Any help (and another approach) is welcome.

Thanks!

Answer 1

Suppose that $G$ isn't cyclic. Let $x\in G$. Then $d=| x|\lt n$. By Euler's formula, $$n=\sum_{d\mid n}\varphi(d)\gt\sum_{d\mid n,d\ne n}\varphi (d)\ge| G|.$$

Contradiction.

Answer 2

I think there is another proof not using the partition of $G$ into $G_d$.

First we prove the theorem for all finite abelian groups. Let $G$ be a finite abelian group, which is decomposed as $G = C_{d_1} \oplus C_{d_2}\oplus\dots\oplus C_{d_m}\,$, $d_1|d_2|\dots|d_m$. If $m\gt1$, then there is a subgroup $D\lt C_{d_2}$ of order $d_1$, contradicting our hypothesis. So $m = 1$, $G=C_{d_1}$ is cyclic.

It remains to show that any finite group satisfying our hypothesis must be abelian. So let $G$ be a finite group. We show first that all Sylow subgroups of $G$ are normal, from which it follows that $G$ must be the direct product of its Sylow subgroups. If $P\in Syl_p(G)$ and $P^g \not= P$ for some $g\in G$, there is some $h \in P$ with $h \notin P^g$. $\langle h \rangle$ and $\langle h^g \rangle$ are two distinct subgroups isomorphic to each other, for otherwise $h \in \langle h^g \rangle \lt P^g$.

We have proven that all Sylow subgroups of $G$ are normal, so $G$ is indeed the direct product of its Sylow subgroups. If we prove that all Sylow subgroups of $G$ are cyclic, then $G$ is abelian, completing the proof. Let $P$ be a group of order $p^n$ satisfying our hypothesis. We prove by induction on $n$. $P$ must have a normal subgroup $Q$ of order $p^{n-1}$, which is already cyclic by the induction hypothesis. Pick an element $a \in P$ with $a \notin Q$. If $a$ has order $p^n$, we are done; otherwise, assume $\langle a\rangle$ has order $p^m$, $1\leq m \lt n$. $Q$, being cyclic, also has a subgroup $\langle b \rangle$ of order $p^m$; $\langle a\rangle \neq \langle b \rangle$ since $a \notin Q$, contradicting the hypothesis. Thus $P$ is cyclic.