Prove $G$ is cyclic if for every $n\in \Bbb{N}$, there are at most $n$ solutions to the equation $x^n=e$.

Question

Let $G$ be a finite group. Prove G is cyclic if for every $n\in \Bbb{N}$, there are at most $n$ solutions to the equation $x^n=e$.

I am stuck with this :( Would appreciate your help.

Answer 1

I think the answer would run along the following lines.

Suppose that $G$ is not cyclic. Take $a\in G$, and take a $b\in G$, such that $b$ cannot be written as $a^n$ for any number $n$ (such a $b$ exists, because $G$ is not cyclic).

Now, let the order of $a$ be $k$. Then $a, a^2, a^3, \ldots, a^{k-1}$ are distinct elements in your group, and they all have the property that $x^k=e$, where $x$ is in that list. Notice also that $b$ is not contained in this list. If the order of $b$ is $n$, then $b, b^2, \ldots, b^{n-1}$ are $n-1$ elements, such that each fulfills $x^n=e$.

Say $k<n$. The question remains if there is some element in the list $a, a^2, \ldots, a^{k-1}$, such that $x^n=e$, and if not, will there be some $b$ for which this is true.