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Is there any set of mathematical objects that satisfies all of the following:

1) For each object $a$ in the set, $a^2$ is some multiples of $x$.

2) $ab$ is never multiples of $x$ where $a$ and $b$ are any different two objects in the set.

3) all objects commute - $ab = ba$ and $a+b = b+a$.

Does such set exists for all cardinality?

Edit: to specify, let objects be matrices.

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What kind of structure? What is $x$? What is the motivation? –  Marc van Leeuwen Nov 30 '12 at 12:32
    
any structure - $x$ is a given one. –  KDE_dolphin Nov 30 '12 at 12:34
    
specified: matrices. –  KDE_dolphin Nov 30 '12 at 13:48
    
When you say "multiples of x", do you just mean scalar multiples ($2x, pi\cdot x$, etc.)? –  anonymous Dec 1 '12 at 11:17

2 Answers 2

Take a set $Y$ of the required cardinality, $x$ an element not in $Y$, and $S$ the algebra of some field $K$ with generators $\{x\}\cup Y$ and relations $a^2=x$ and $ab=ba$ for all $a,b\in S$.

Seriously, if you don't want to specify what your question is about, you can only expect trivial answers like this.

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Did you mean to write "with generators $\{x\}\cup Y$"? –  Gregor Bruns Nov 30 '12 at 13:12
    
@GregorBruns: yes, thank you. –  Marc van Leeuwen Dec 1 '12 at 10:38

Let $S$ be your set with the proposed structures, and $x$ is the distinguished element. I can only assume that you require $x \in S$, and that $S$ is closed under the usual operations of $+$ and $\times$, since otherwise, property (1) makes little sense.

Your property (2) implies that your set $S$ can only consist of $\{x\}$ (a singleton set). Indeed, if $a \neq x$, then $a\cdot x$ should not be a multiple of $x$, but certainly $ax$ is a multiple of $x$. Thus $a=x$.

Thus, only for cardinality = $1$ will $S$ exist.

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