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, $$n=\sum_{d\mid n}\varphi(d)\gt\sum_{d\mid n,d\ne n}\varphi (d)=| G|.$$

The reason we can use Euler's formula here is that by hypothesis there's at most one cyclic subgroup of each order dividing $n$.