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?
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asked Feb 19, 2023 at 19:18
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3$\begingroup$ $1+1 = 0$ in $\mathbb Z_2$ $\endgroup$– Brevan EllefsenCommented Feb 19, 2023 at 19:25
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$\begingroup$ @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. $\endgroup$– Antonio Maria Di MauroCommented Feb 19, 2023 at 19:32
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2$\begingroup$ $\Bbb Z_2$ 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 $a^2 = e$. $\endgroup$– MJDCommented Feb 19, 2023 at 19:45
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6$\begingroup$ 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? $\endgroup$– user700480Commented Feb 19, 2023 at 19:50
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1$\begingroup$ @PseudoNeo Oh, really nice theorem!!! Well, unfortunately I did not know it before you enunciated here. $\endgroup$– Antonio Maria Di MauroCommented Feb 19, 2023 at 21:55
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2 Answers
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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?
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$\begingroup$ 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. $\endgroup$ Commented Feb 20, 2023 at 9:45
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1$\begingroup$ 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 $1$. $\endgroup$– user700480Commented Feb 20, 2023 at 10:21
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$\begingroup$ Okay, all clear: answer approved. Thanks very much for your assistance! $\endgroup$ Commented Feb 20, 2023 at 11:01
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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 $1+1=0$.
