# Solve $x^n=1$ in a monoid.

Let $$(X, *)$$ a monoid with identity $$e$$. So can the equality $$x^n=e$$ hold for some $$n\ge 1$$ when $$x$$ is not equal to $$e$$? If this can be true what is an example? If this is not true how prove it?

• $1+1 = 0$ in $\mathbb Z_2$ Commented Feb 19, 2023 at 19:25
• @BrevanEllefsen Sorry my ignorance but I started to study Algebra seriously only today: what is $\Bbb Z_2$? I point out by Linear Algebra I know what is a group, what is a ring, what is a Field. Commented Feb 19, 2023 at 19:32
• $\Bbb Z_2$ is the monoid where $X=\{e, a\}$ and $\ast$ is the function that has $e\ast e = a\ast a = e$ and $e\ast a = a\ast e = a$. In this monoid $x^n=e$ has the solution $a^2 = e$.
– MJD
Commented Feb 19, 2023 at 19:45
• Just take the set of all complex numbers on the unit circle: $U=\{z\in\mathbb C\mid |z|=1\}$, with ordinary multiplication. This is a monoid. (Right?) The role of $e$ is played by $1$. Do equations $z^n=1$ have solutions other than $z=1$? Now take $(0, +\infty)\subseteq \mathbb R$ ("positive real numbers") with ordinary multiplication. This is a monoid and $1$ is again in the role of $e$. Does $x^n=1$ imply $x=1$ there?
– user700480
Commented Feb 19, 2023 at 19:50
• @PseudoNeo Oh, really nice theorem!!! Well, unfortunately I did not know it before you enunciated here. Commented Feb 19, 2023 at 21:55

Just take the set of all complex numbers on the unit circle: $$U=\{z\in\mathbb C\mid|z|=1\}$$, with ordinary multiplication. This is a monoid. (Right?) The role of $$e$$ is played by $$1$$. Do equations $$z^n=1$$ have solutions other than $$z=1$$?
Now take $$(0,+\infty)\subset\mathbb R$$ ("positive real numbers") with ordinary multiplication. This is a monoid and $$1$$ is again in the role of $$e$$. Does $$x^n=1$$ imply $$x=1$$ there?
• To complete the answer: into the first case I would put $z$ equal to $i$ and $n$ equal to $4$ but for the second case I suppose that only $1$ solve the equation. Anyway, thanks for the answer: I just upvoted it and if you reply to my comment adding more details about the second case (are there any other observations to add to the ones I just made?) then I surely will accept it. Commented Feb 20, 2023 at 9:45
• In the first case you can put $z$ equal to $\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$ and then it will work for any $n$ (De Moivre's formula.) In the second case, on $(0, +\infty)$ the function $x\mapsto x^n$ is strictly increasing so it can have any value at most once. As $1^n=1$, no other value $x$ can map to $1$.
Since every abelian group is equipped with a commutative multiplication and an identity, it suffices to consider a finite abelian group. Just as Brevan has mentioned, we can consider $$\mathbb{Z}/2\mathbb{Z}$$, or the cyclic group of order 2 which contains the identity and only one non-trivial element $$1$$. Since this group is cyclic we must have $$1+1=0$$.