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In a finite group $G$ with $e:=\text{id}_G$, show that the number of nonidentity elements that satisfy the equation $x^5=e$ is a multiple of 4.

This is number $50$, Ch.$2$ from Gallian's text. I have seen two repeats of this question on MSE (Show that number of solutions satisfying $x^5=e$ is a multiple of 4?, In a finite group, show that the number of nonidentity elements that satisfy the equation $x^5=e$ is a multiple of 4.) but I still have questions about the question and my proof.

My first question:

Do we have to assume that $x^5=e$ for some $x\in G$ in the first place? My thinking is that this is a yes since in general a finite group may not have such an $x$.

Proof of claim:

Suppose some $x\in G$,$\space$$x\neq e$ satisfies the condition $x^5=e$. Then note that $x^2\in G$ and $$(x^2)^5=(x^5)^2=e^2=e$$ so $x^2$ satisfies the condition. Similarly $x^3\in G$ and $x^4\in G$. Observe that $$(x^3)^5=(x^5)^3=e^3=e$$ $$(x^4)^5=(x^5)^4=e^4=e$$ so that $x^3$ and $x^4$ also satisfy the condition. Once we verify that $x,x^2,x^3,x^4$ are distinct and that $x^i\neq e$ for $1\leq i \leq 4$ we will have proved the claim since for every $x$ that is a solution, so is $x^2$,$x^3$, and $x^4$. Thus solutions come in multiples of $4$.

Note that we don't consider elements like $x^6$ or $x^7$ as solutions because $x^6=x$ and $x^7=x^2$ i.e. for $n>5$, $x^n=x^i$ where $i\in\{1,2,3,4,5\}$. So, considering powers of elements modulo $5$ is enough.

To show each $x^i$ is distinct, we assume the contrary. That is, assume $$x^i=x^j$$ for some distinct $i,j\in\{1,2,3,4\}$. Thus if we take $i$ to always be the greater of the two, $$x^i=x^j\iff x^{i-j}=e$$ so $$i-j=1,2\text{ or } 3$$

Note it is impossible $i-j=1$ by the assumption that $x\neq e$. If $i-j=2$ or $i-j=3$, then $$x^2=e\text{ and } x^3=e\implies x^3=x^2\cdot x=e\cdot x=x = e$$

but the latter shows $x=e$ if $x^3=e$ and $x^2=e$; this is a contradiciton. Thus it must be false that $x^i=x^j$ for some distinct $i,j\in\{1,2,3,4,5\}$.

In proving the above, we saw that the $x^2=e=x^3$ leads to a contradiction, so to show the last claim, namely that $x^i\neq e$, we show that $x^4\neq e$. Again assume that indeed $x^4=e$. Then $$x^4=e=x^5\iff e=x\therefore\text{ contradiction }$$ $\blacksquare$

My second question:

Is the above proof correct? By removing the condition that the group be finite, how could this change the conclusion about the number of solutions? I never really utilized that $G$ was finite above (maybe tacitly? I don't know) so I'm pretty stumped on this one.

Arturo Magidin
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    For your first point, you do not have to assume that $x^5 = e$ for some $x\neq e$, since if this doesn't hold, the number of such elements is 0 and 4 does divide 0. For why you assume finiteness, note that the infinite product of $\mathbb{Z}_5$ with itself has an $infinite$ number of such elements, and so the claim that 4 divides this number does not make sense. Basically, the finiteness of the group guarantees the finiteness of the set under consideration. Proof looks good! – Andrew Sep 06 '19 at 04:47
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    You still need to show that no two sets of four elements you make in this way have an element in common (hint: if you think in terms of subgroups and Lagrange's thm, this is an easy observation) – Brian Moehring Sep 06 '19 at 04:50
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    It is not correct for me. For example if you consider x such that $x^2=e$, then you have that it satisfies your condition but $x^3=x$ – Federico Fallucca Sep 06 '19 at 04:52
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    @Federico If $\exists x$ s.t. $x^2=e$ then $x=x^{-1}$ so $x^5=x\cdot x\cdot x\cdot x\cdot x=x\cdot x^{-1}\cdot x\cdot x^{-1}\cdot x=e\cdot e \cdot x=x$. So that $x$ wouldn't satisfy the condition, right? –  Sep 06 '19 at 04:58
  • @BrianMoehring I saw something like that from James in the first answer I linked... although by chapter 2, Gallian has only developed uniqueness of identity and inverse, cancellation, and the 'shoe-socks' property. To show what you're proposing, take $x$ and $y$ as solutions. Then we have to show ${x,x^2,x^3,x^4}\cap{y,y^2,y^3,y^4}=\emptyset$? –  Sep 06 '19 at 05:08
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    Almost. The intersection should either be empty or equal to ${x,x^2,x^3,x^4}$ The easiest way, absent lagrange's theorem, is to first assume $y=x^i$ and show you get the whole set again. Then assume $y$ isn't one of the powers of $x$ but $x^i=y^j$ and reach a contradiction. – Brian Moehring Sep 06 '19 at 05:30
  • Okay @BrianMoehring, I will try that. Thank you! –  Sep 06 '19 at 05:36

2 Answers2

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In general, if $p$ is an odd prime and $G$ a finite group, then $\#\{ g \in G: g^p=1\} \equiv 1$ mod $(p-1)$. Observe that the set includes the identity element. Proof (sketch): on the set $S=\{ g \in G: g^p=1\}$ define an equivalence relation: $g \sim h$ if and only if $\langle g \rangle =\langle h \rangle$. Then $S$ partitions in $\{1\}$ and equivalence classes of order $p-1$ (namely $\langle g \rangle -\{1\}$ for each non-identity $g \in S$).

Nicky Hekster
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Let $G$ be a finite group with identity $e$. Let $x\in G$ such that $x^5=e$. Then as you obtained there are four other elements that follows same relation. Your proof is correct, but you make this so cumbersome. By the way I am giving a general idea to you, for any $x\in G$ such that $x^k=e$ the number of non-identity elements of $G$ satisfy the equation $x^k=e$ is a multiple of $\phi(k)$, where $\phi$ is Euler function. You make see this in an easy way, using the formula,

$$\circ(x^r)=\frac{\circ(x)}{gcd(r,\circ(x))}$$

MANI
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