I have trouble proving the following statement:
Suppose that $H$ is a finite group with order $n$ and $e$ is the identity element of $H$. For an arbitrary positive integer $d$ satisfing $d\mid n$, the set $\{ x \in H : x^d = e \}$ has at most $d$ elements. Then we can make a conclusion that $H$ is a cyclic group.
Here is my idea:
First we need one lemma $($ notation $|y|$ denotes the order of $y$ $)$ :
Lemma 1 Let $G$ be a commutative group. If each element $x \in G$ satisfies $|x| \leq m$ $($ there exists an element $y \in G$ sucht that $|y|=m$ $)$, then $G = \{ x \in G : x^m = e \}$.
To prove it, we only need to prove that there exists an element $y \in H$ such that the order of $y$ is $n$. To begin with, we first assume that $H$ is commutative ( I can't prove this ). Next, we suppose $n_1 < n$ and $n_1$ is the biggest factor of $n$. Obviously, we have $A \triangleq H- \{ x \in H : x^{n_1} = e \} \neq \emptyset$. If $|y| < n_1$ for any $y \in A$, then by Lemma 1 we come to a conclusion that $H = \{ x \in H : x^{n_1}=e \}$. It is a controdiction to the fact that the set $\{ x \in H : x^{n_1} = e \}$ has at most $n_1$ elements. Hence, there exists an element $y \in A$ such that $|y| > n_1$. Since $|y| \mid n$, we get $|y|=n$. Therefore, this completes the proof.
My question:
The only thing I have not done is to prove the group $H$ is commutative. Can anybody help me ?
Supplement:
The proof of Lemma 1: Let $a \in G$ and $|a|=m$. Suppose that $b$ is an arbitrary element in $G$ with order $|b|=n$. If $n \nmid m$, there must exist a prime $p$ such that $$ m=p^k m_1 , p \nmid m_1 , $$ $$ n=p^tn_1 , t>k. $$ Since $|a|=m$, $|b|=n$, we have $|a^{p^k}|=m_1$, $|b^{n_{1}}|=p^t$. By the property $( m_1, p^t )=1$ and the fact that $G$ is a commutative group, we get $$ |a^{p^k}b^{n_1}|=p^t m_1 > p^k m_1 = m. $$ This is a contradiction.