# A question on cyclic group with finite order

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.

• H does not have to be commutative for the statement to hold. A group is cyclic iff it has a generator (an element of order n). Aug 23, 2014 at 17:13
• @Bungo Actually we were both wrong: $C_2 \times C_3 \simeq C_6$. Aug 23, 2014 at 17:15
• @angryavian: LOL, need more coffee :-)
– user169852
Aug 23, 2014 at 17:15
• The lemma is from a book. I'll prove the Lemma later. Aug 23, 2014 at 17:17
• @forallepsilon: Cyclic group is commutative. I can't prove the even weaker case. Aug 23, 2014 at 17:48

To show a finite group is cyclic, a good first step is to show nilpotency, a weakened form of abelianness (indeed, a finite group is nilpotent iff $ab = ba$ whenever $|a|$, $|b|$ are coprime). This allows one to reduce to the case of $p$-groups, which is often simpler.
To this end, let $p \mid |G|$, and let $P$ be a Sylow $p$-subgroup of $G$, say $|P| = p^n$. Then by assumption, $G$ has at most $p^n$ elements of order a power of $p$, so $P$ is the unique Sylow $p$-subgroup of $G$. Thus every Sylow $p$-subgroup of $G$ is normal $\iff G$ is nilpotent $\iff G \cong P_1 \times \ldots \times P_n$, where $P_i$ are the Sylow subgroups of $G$ (for different primes).
It remains to show that each Sylow subgroup $P$ of $G$ is cyclic. If $|P| = p^n$, set $S := \{x \in P \mid x^{p^{n-1}} = e\}$. If $P$ were not cyclic, then for every $y \in P$, $|y| \mid p^n$ but $|y| \ne p^n$, so $|y| \mid p^{n-1} \implies y^{p^{n-1}} = e \implies y \in S$. This implies $P \subseteq S$, but by assumption $|S| \le p^{n-1}$.
Thus $G$ is a direct product of its Sylow $p$-subgroups, which are each cyclic and of relatively prime order, so $G$ is cyclic.