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I am trying to prove problem 48 of chapter 2 in the book Contemporary Abstract Algebra by Gallian. I saw solutions on math stack exchange

In a finite group, show that the number of nonidentity elements that satisfy the equation $x^5=e$ is a multiple of 4.

Show that number of solutions satisfying $x^5=e$ is a multiple of 4?

But I didn't understand completely. Here is my understanding:

To prove the statement first we start from assuming that $x=a$ a non-identity element satisfy the equation $x^5=e$, then we find three more solutions. After that we shall show all four solutions are not identity and distinct. At the end we conclude our proof by showing any other element out of the list, of four solutions, that satisfy equation will also belong to the list thus the four solutions are actually all solutions of equation.

Proof Suppose $a$ is a non-identity solution of $x^5=e$. Then we see $x=a^i$ for $1<i≤4$ is also solution of $x^5=e$, since $(a^i)^5=(a^5)^i=e$. Thus {$a,a^2,a^3,a^4$} is the set of element that satisfy $x^5=e$.

Suppose to the contrary that any two elements of the set {$a,a^2,a^3,a^4$ } are equal say $a^i=a^j=e$ for $i≠j$. Then $a^{i-j}=e$ thus $i-j$ is equal to 2 or 3, so $a^2=e=a^3$ which implies $a=e$. A contradiction.

We see each element of the set {$a,a^2,a^3,a^4$ } is non-identity. Since if $x^4=e$ and $x^5=e$, then $x=e$, using same reasoning we conclude all are non-identity.

Let $x=a^n$ is a solution of the equation and doesn't belong to the above set, that means $n≥5$. By division lemma $a^n = a^{5q+r}=(a^5)^qa^r= a^r=e$ where $1≤r≤4$. But $a^r \in$ {$ a, a^2,a^3,a^4$ }. A contradiction. Notice $r=0$ gives identity solution so we exclude it.

This shows if any element $x$ satisfies the equation, then it must belong into the list and hence solutions are multiple of 4 because if $x$ belongs to the set, then $x^2,x^3,x^4$ will also belong. QED

My Questions

  1. My thinking about the proof and proof are correct?
  2. What is the standard example of above statement? I tried to find example but i found trivial example in which there was 0 solutions to the equation (i.e. multiple of 4).
  3. I don't understand what will happened if the group is not finite.
stoic-santiago
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Afzal
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2 Answers2

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Let's deal first with question 3. If $G$ is an infinite group, it may have infinitely many elements satisfying $x^5=e$, and it makes no sense to ask whether infinity is a multiple of $4$.

"At the end we conclude our proof by showing any other element out of the list, of four solutions, that satisfy equation will also belong to the list thus the four solutions are actually all solutions of equation." No, it is not necessarily the case that there are only four non-identity elements satisfying $x^5=e$. The number is a multiple of four, but it could be any multiple of four (and that includes zero). Just for an example, consider the group $S_5$, the group of all permutations of five letters. Every five-cycle in $S_5$ has $x^5=e$, and there are $24$ of these.

What you need to do is
prove that if $a$ is a non-identity solution of $x^5=e$, then so are $a^2,a^3,a^4$, and these are distinct, and
prove that if $b\ne a$ is another non-identity solution of $x^5=e$, and if some power of $b$ is a non-identity power of $a$, then every power of $b$ is a power of $a$. This insures that the four elements $b,b^2,b^3,b^4$ are either identical (in some order) to $a,a^2,a^3,a^4$, or else the two sets of four are completely different.

Gerry Myerson
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For $p$ a prime, if $x$ is nontrivial and $x^p=e$, then $\langle x\rangle$ is a subgroup of order $p$, and each of the $p-1$ nontrivial elements $y\in\langle x\rangle$ fulfils the equation $y^p=e$. If $z\notin\langle x\rangle$ also yields $z^p=e$, since $\langle x\rangle\cap \langle z\rangle$ is trivial, then there are other $p-1$ nontrivial elements $k\in\langle z\rangle$ such that $k^p=e$, etc. Therefore, the nontrivial elements $w$ such that $w^p=e$ come in multiple of $p-1$.

citadel
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