$\aleph_0+\aleph_0=\aleph_0$? or $2\aleph_0$?

Question

The problem is 'In a finite group, show that the number of nonidentity elements that satisfy the equation $x^5=e$ is a multiple of 4.'. I did it and got answer that cause $x$, $x^2$, $x^3$, $x^4$ always come together but when the condition that group be finite is omitted, what can I say the number of nonidentity elements that satisfy the equation? I first thought it would be $\aleph_0+\aleph_0+\aleph_0+\aleph_0$. But $\aleph_0+\aleph_0=\aleph_0$? or $2\aleph_0$? This algebra problem makes me solve set theory problem.

Answer 1

As you've noted if $g^5=e$ for some $g\in G$ then $x^5=e$ for any $x\in R_g=\{g,g^2,g^3,g^4\}$ and these sets are either equal or disjoint, and $|R_g|=4$ if $g\neq e$. And so if $k$ is the number of nontrivial solutions to $x^5=e$ and $k<\infty$ then indeed $4$ divides $k$. Note that I didn't assume that $G$ is finite, only that $k$ is. Also note that this can be generalized to any prime number: for any prime $p$ the number of (nontrivial) solutions to $x^p=e$ is divisible by $p-1$ if finite. For non-primes this doesn't hold since different $R_g$ sets need not be disjoint (and so they don't partition all solutions) and need not be of equal size.

Now if $\mathcal{K}$ is any infinite cardinal number then I will show that there is a group $G$ having exactly $\mathcal{K}$ solutions to $x^5=e$, assuming the Axiom of Choice. And so your conclusion "there is $\aleph_0+\aleph_0+\aleph_0+\aleph_0$ solutions" is incorrect (even though $\aleph_0=2\aleph_0$ which is not really relevant).

Let $G$ be the infinite direct sum $\bigoplus_{\mathcal{K}}\mathbb{Z}_5$, which has exactly $\mathcal{K}$ elements by the AoC. And so $G$ has at most $\mathcal{K}$ solutions to $x^5=e$.

On the other hand $G$ has at least $\mathcal{K}$ solutions to $x^5=e$, namely for a fixed $i\in \mathcal{K}$ (here I treat $\mathcal{K}$ as a set of size $\mathcal{K}$) we have

$$f_i:\mathcal{K}\to\mathbb{Z}_5$$ $$f_i(x)=\begin{cases}1 &\text{if }x=i \\ 0 &\text{otherwise} \end{cases}$$

which is a solution and $f_i=f_j$ if and only if $i=j$. And so there are at least $\mathcal{K}$ solutions to $x^5=e$.

All in all $G$ has exactly $\mathcal{K}$ solutions to $x^5=e$.

Finally note that an infinite group $G$ can have any (valid) finite number of solutions to $x^5=e$ as well. For example if $G$ is finite with $k<\infty$ solutions then $G\times\mathbb{Z}$ is infinite with $k$ solutions.

//Edit: As @bof noticed the Axiom of Choice is not needed for $2^{\aleph_0}$ case since the group of rotations of a two-dimensional sphere has $2^{\aleph_0}$ elements of order $5$ (or any other order $\geq 2$). These are rotations around different lines in $\mathbb{R}^3$ going through the origin.

Answer 2

Both are correct, because $\aleph_0 = 2\aleph_0$

For example, $|\mathbb{N}|=\aleph_0$ and $|\mathbb{Z}| = \aleph_0$ as well. I think this illustrates why that is true.

Answer 3

Function $f\colon\mathbb N\sqcup \mathbb N \to \mathbb N$ defined by $f(n,i) = 2n + i$ is a bijection and thus $\aleph_0+\aleph_0 = |\mathbb N\sqcup \mathbb N| = |\mathbb N| = \aleph_0.$

Answer 4

By Hessenberg theorem $\kappa\cdot\kappa=\kappa$, for $\kappa$ infinite cardinal This implies $\kappa\leq 2\kappa=\kappa+\kappa\leq \kappa\cdot\kappa=\kappa$. So, $\kappa=2\kappa=\kappa+\kappa=\kappa\cdot\kappa$

So in particular $\aleph_0=\aleph_0+\aleph_0=2\aleph_0$. So I guess the answer to your question is both.