Finite group for which $|\{x:x^m=e\}|\leq m$ for all $m$ is cyclic. [duplicate]

Solution 1:

Yes, that is perfect. Let $|G|=n$. Note that

$$G=\bigsqcup_{d\mid n}X_d$$

where $X_d$ is the set of elements of $G$ of order $d$. Now, if we can show that our restriction requires $\#(X_d)\leqslant \phi(d)$ then the equality $\displaystyle n=\sum_{d\mid n}\phi(d)$ will actually force $\#(X_d)=\phi(d)$, for all $d\mid n$, and so, in particular $\#(X_n)>0$.

Now, suppose that there were more than $\phi(d)$ elements of $G$ of order $d$. Note then that since the cyclic group $\langle x\rangle$, for any $x\in X_d$, has exactly $\phi(d)$ elements of order $d$, there must exist another element $y\in G$ with $|y|=d$ and $y\notin \langle x\rangle$. But, Lagrange's theorem then implies that we've produced $|\langle x\rangle|+1=d+1$ solutions to $x^d=1$--contradictory to assumption.

Solution 2:

Here's a different proof that uses Sylow's theorems. Suppose that $x^m = e$ has at most $m$ solutions in $G$ for all positive integers $m$.

Then for every prime $p$ dividing $|G|$, there exists at most one Sylow $p$-subgroup. Otherwise we would find more than $p^{\alpha}$ solutions to $x^{p^{\alpha}} = e$, where $p^\alpha$ is the largest power of $p$ dividing $|G|$. The Sylow $p$-subgroup is cyclic, since $x^{p^{\alpha-1}} = e$ has less than $p^\alpha$ solutions, so there exists an element $g$ satisfying $g^{p^\alpha} = e$ and $g^{p^{\alpha-1}} \neq e$.

Because every Sylow $p$-subgroup of $G$ is normal and cyclic, the group $G$ is cyclic.

There is a stronger statement due to Cohn, see the following paper (PDF link).

J. H. E. Cohn, A condition for a finite group to be cyclic, Proc. Amer. Math. Soc., Vol. 32, No. 1 (1972).

In this article (it is short, and the proof is simple counting argument) Cohn proves that a finite group $G$ is cyclic if for every prime power $p^k$, the equation $x^{p^k} = e$ has at most $p^{k+1} - 1$ solutions.