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I am searching for examples of groups which contain their own operation as an element. I am having difficulty showing that this is not possible for groups of size greater than 1, but counterexamples are also elusive.

Taking the set $S=\{f\}$
If we define
$f:(S \times S) \to S$
$f(f,f)=f$
Then we can construct a group $G$ on the set $S$ with operation $f$. The identity of $G$ is $f$, and this group satisfies all required properties.

My question is: Are there groups of size greater than 1 which contain their operation as an element?

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I don't understand your example, your definitions look circular to me since you use f in the definition of S and then S in the definition of f! –  Noah Snyder Oct 5 '12 at 16:11
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1 Answer

up vote 1 down vote accepted

To think about this at all, you'll have to be working in an ill-founded set theory which allows sets to contain themselves: in particular the normal set theoretic description of your $f$ as $f={(f,f)}$ contradicts the Axiom of Foundation of standard Zermelo-Fraenkel set theory.

You'll have a similar problem trying to include a group's operation as an element in a larger group. If I have some group $(G,f), f:G\to G,$ then $f$ contains an element $(f,g)=\{f,\{f,g\}\}$ which contains $f$ which contains an element which contains $f$ which...it's an infinite regress.

So the answer within the most common set-theoretic framework is that no such group exists.

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Well, @mike4ty4, your construction might have produced some interesting things in Barwise and Aczel's non-well-founded set theory! Noah, if you're interested in carrying this idea beyond the confines of the consensus axioms, you should look up those names. It's a small enough field that conceivably no one's ever done much regarding this question. –  Kevin Carlson Oct 2 '12 at 1:40
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