If $x^m=e$ has at most $m$ solutions for any $m\in \mathbb{N}$, then $G$ is cyclic

Fraleigh(7th ed) Sec10, Ex47. Let $G$ be a finite group. Show that if for each positive integer $m$ the number of solutions $x$ of the equation $x^m=e$ in $G$ is at most $m$, then $G$ is cyclic.

I tried it a few hours but I couldn’t solve it. So I read the solution. But I narrowly understood the solution, and it’s still unclear to me. How can I solve it?

Solutions Collecting From Web of "If $x^m=e$ has at most $m$ solutions for any $m\in \mathbb{N}$, then $G$ is cyclic"

The proof that I know of this fact goes as follows. Say that $|G|=n$ and for $1\leq d\leq n$ let
$$
A_d=\{\hbox{$g\in G$ such that $g$ has exact order $d$}\}
$$
Then one proves:
(1) that $A_d=\emptyset$ if $d$ is not a divisor of $n$;
(2) that if $d\mid n$, then $|A_d|\leq\varphi(d)$.

This allows to conclude because the chain of inequalities
$n=|G|=\sum_{d\mid n}|A_d|\leq\sum_{d\mid n}\varphi(d)=n$ gives actually equality at every step and in particular $A_n\neq\emptyset$, i.e. $G$ admits a generator.

There are many ways to proceed with this problem, so I give one which I don’t remember seeing in a text, though perhaps it uses a little more knowledge of group theory. We proceed by induction. The result is true if $|G| =p$ for some prime $p$, (and if $|G| =1$), so suppose that $|G| >1$ is not prime and the result is true for groups of order less than $|G|$. If $H$ is any proper subgroup of $G$, then $H$ is cyclic. If $H$ has order $m$, then $H$ contains $m$ solutions in $G$ of $x^m = 1$. But there are only m solutions of $x^m = 1$ in $G$, so $H = \{ x \in G: x^m = 1 \}$. Notice that if $x^m = 1$
then $(gxg^{-1})^m = 1$ for any $g \in G$. Hence $gHg^{-1} = H$ for any $g \in G$.
Hence every subgroup of $G$ is normal. In particular, each Sylow $p$-subgroup of $G$
is normal in $G$, and $G$ is the direct product of its Sylow $p$-subgroups. If all Sylow
$p$-subgroups of $G$ are proper, then all are cyclic by the induction hypothesis, and then $G$ itself is cyclic. Hence we may suppose that $G$ is a $p$-group for some prime $p$. Let $p^e$ be the largest order of an element of $G$. Then $G$ has a cyclic subgroup, say $N$, of order $p^e$, and $N \lhd G$ as above. But choose $x \in G$.
Then $x$ has order $p^f$ for some non-negative integer $f$. If $f >e$, the choice of $e$
is contradicted. If $f \leq e$, then $x \in N$, since $x^{p^e} = 1$ and $N$ contains
all solutions in $G$ of $y^{p^e} = 1$. Hence $N = G$ and $G$ is cyclic.

In addition, there is a very interesting theorem of Frobenius that says that if G is a finite group, $n$ a positive integer dividing the order $|G|$, then $n$ divides $|\{ x \in G: x^n = 1 \}|$. See for example The American Mathematical Monthly, Vol. 99, No. 4, Apr., 1992.

Please refer I.N. Herstein ” Topics in Algebra” Second Edition. Page $358$. Chapter: Selected Topics.

He solves it by considering different cases. The solution given in his book is as follows:

If the order of $G$ is a power of some prime number $q$, then the result is very easy. For suppose that $a \in G$ is an element whose order is as large as possible: its order must be $q^{r}$ for some integer $r$. The elements $e,a, a^{2},\cdots, a^{q^r-1}$ give us $q^{r}$ distinct solutions of the equation $x^{q^r}=e$, which by our hypothesis implies that these are all the solutions of this equation. Now if $b \in G$, its order is $q^{r}$ where $s \leq r$, hence $$b^{q^r} = (b^{q^s})^{q^{r-s}}=e$$ By the observation made above this forces $b=a^{i}$ for some $i$ and so $G$ is cyclic.

The general finite abelian group $G$ can be realized as $G=S_{q_1} \cdot S_{q_2} \cdots S_{q_k}$ where the $q_i$ are the distinct prime divisors of $|G|$ and where the $S_{q_i}$ are the sylow subgroups of $G$. Moreover, every element of $g \in G$ can be written in a unique way as $g=s_{1}s_{2}\cdots s_{k}$ where $s_{i} \in S_{q_i}$. An solution of $x^{n}=e$ in $S_{q_i}$ is one of $x^{n}=e$ in $G$ so that each $S_{q_i}$ inherits the hypothesis we have imposed on $G$. By the remarks of the first paragraph of the proof, each $S_{q_i}$ is a cyclic group; let $a_i$ be a generator of $S_{q_i}$. We claim that $$c=a_{1}\cdot a_{2} \cdots a_{k}$$ is a cyclic generator of $G$. To verify this, all we must do is to prove, that $|G|$ divides $m$, the order of $c$. Since $c^{m}=e$, we have that $a_{1}^{m}\cdot a_{2}^{m} \cdots a_{k}^{m}=e$. By the uniqueness of representation of an element of $G$, as a product of elements in the $S_{q_i}$ we conclude that each $a_{1}^{m}=e$. Thus $|S_{q_i}| \mid m$ for every $i$. Thus $$|G| = |S_{q_1}| \cdot |S_{q_2}| \cdots |S_{q_k}| \ \Bigl|\: m$$ However $m \mid |G|$ and so $|G|=m$. This proves that $G$ is cyclic.

Hint: First of all, it is enough to have the property for each divisor of $|G|.$ The main ingredient is the following fact for every positive integer $n$, that $n=\sum_{d|n} \varphi(d)$, for all divisors $d$ of $n$, where $\varphi$ is the Euler’s function.

Here’s another method without using Euler’s totient function or induction. We rely on a lemma:

If $H$ and $K$ are normal subgroups of $G$ such that $H\cap K=\{e\}$, then $$\forall x\in H\;\forall y\in K(xy=yx)$$

Let $G$ satisfy the above conditions, and let’s denote its order by $k$. Then since $G$ is finite, we can find an element of maximal order. Let’s pick one of these elements and call it $x$, and let’s call its order $n$. If $n=k$, we would be finished as then $G=\langle x\rangle$.

Suppose for contradiction that $k\neq n$. Then there is an element $y\in G\setminus \langle x\rangle$. Let’s denote $y$’s order by $m$, and let’s choose $y$ to have minimal order among the elements of $G\setminus\langle x\rangle$. By counting arguments, we have that $\langle x\rangle=\{z\in G: z^n=e\}$. We will use $x$ and $y$ to create a new element $z$ which satisfies $z^n=e$ which is not in $\langle x\rangle$ contradicting our assumption.

Consider a conjugate subgroup of $\langle x\rangle$. Such a subgroup would also be cyclic of order $n$, and thus have elements which solve $z^n=e$. Thus there cannot be a distinct conjugate subgroup of $\langle x\rangle$ else we would have too many elements satisfying $z^n=e$. Likewise, $\langle y\rangle$ has no non-trivial conjugate subgroups. This means that $\langle x\rangle$ and $\langle y\rangle$ are normal. Now we consider two subcases:

Case 1: $\gcd(m,n)=1$

In this case, we have that $\langle x\rangle\cap\langle y\rangle=\{e\}$. Thus, we have that all powers of $x$ and $y$ commute. And looking at the element $xy$ gives us an element of order $\text{lcm}(m,n)$. Since in general, $\text{lcm}(m,n)\geq n$ for any $m$, we can conclude $\text{lcm}(m,n)=n$ since $n$ was the maximal order. Further, $xy\not\in\langle x\rangle$ for otherwise that would imply that $y\in\langle x\rangle$. Thus there is another solution to $z^n=e$. A contradiction.

Case 2: $\gcd(m,n)>1$.

Let $d=\gcd(m,n)$. Consider the element $y^{m/d}$. It has order $d$. In general $d=\gcd(m,n)\leq m$. So that $y^{m/d}$ has order less than or equal to $m$. Since $m$ was chosen so as to be minimal, we have $m=d$. Thus we have that $m\mid n$. But this means that $y^n=e$. And thus there is another solution to $z^n=1$. A contradiction.

Thus we must have that $k=n$. That is, $G$ is cyclic.