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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?

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If you describe the provided solution stating what parts you don't understand, I'll be glad to fill in the details for you. – user3533 Jul 11 '11 at 8:00
@user3533 Actually, I understood the solution. But It doesn't come to my mind and I'm not satisfied with it. I want to get some other extrapositions. – Gobi Jul 11 '11 at 8:51
-1: You haven't explained what you find unsatisfactory about the given solution. So a lot of people have typed up solutions, which may or may not be the same, or for which you have the same issue, as the solution you've already read. For that matter, a proof is given on p. 3 of But maybe it's the same as the one you already have: I don't own the book you are referring to, so how should I know? You could, at least, show us the solution that you didn't like. – Pete L. Clark Jul 11 '11 at 9:28
@Pete: In my defense, I can't exactly figure out why I disliked the solution. It's just somewhat wordy(as opposed to Mori's mathematical symbolitic expostion) and unclear. Even if some solutions uses the same idea, the detailed description can be different. Also my lack of fluency in English makes me hard to explain the question exactly. But you're right and I'm sorry that I have missed the solution. I will make question more clear next time. – Gobi Jul 12 '11 at 2:31
Is there a proof for this statement via the contrapositive? (Show that in every finite non-cyclic group, $x^m=e$ has more than $m$ solutions for some positive integer $m$.) – Lynn May 16 '15 at 3:32
up vote 14 down vote accepted

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.

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@Jonas Meyer: thanks for editing my answer. Apparently I forgot how to insert braces (I mean, the usual TeX way doesn't work here) – Andrea Mori Jul 11 '11 at 8:50
@Andrea_Mori: Actually, the usual TeX way is also \{ and \}. If you use just { and } they are only grouping constructs; you won't see braces in the output. – ShreevatsaR Jul 11 '11 at 13:34
@ShreevatsaR: Andrea did have \{ and \}. For some reason in this case, \\{ and \\} were required. Andrea Mori: You're welcome. – Jonas Meyer Jul 11 '11 at 17:49
This is the first interesting use of the identity $\sum_{d\mid n}\phi(d)=n$ I have seen. +1 for giving me new appreciation – Bryan Jul 22 '14 at 18:21
@Bryan, that identity is essentially known as Moebius inversion and it is extremely useful! – Mariano Suárez-Alvarez Jul 22 '14 at 20:28

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.

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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.

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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.

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Pardon the pun, but "I am not abel" is a very funny typo for a mathematician :-) – Andrea Mori Jul 11 '11 at 8:55
@Andrea: Sorry, it should i am not able. – user9413 Jul 11 '11 at 10:30
This solution seems to assume that $G$ is abelian (or at least nilpotent, i.e., the direct product of its Sylow subgroups). – Pete L. Clark Jul 15 '11 at 10:33

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.

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Plus $n=\sum_{d|n} \varphi(d)$ is easy to prove by considering all fractions $k/n$ with $1\le k \le n$ and their reduced forms. – lhf Jul 11 '11 at 11:05

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.

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