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Possible Duplicate:
Does every non-empty set admit a group structure (in ZF)?Does every non-empty set admit a group structure (in ZF)?

Let $X$ be an arbitrary nonempty set. Can you define a multiplication making it into an abelian group?

If $X$ is finite, say $|X|=n$, we can just use $X \cong \mathbb{Z}/n\mathbb{Z}$. What if $X$ is infinite?

If I'm not mistaken, the group of permutations of $X$ with finite support has the same cardinality as $X$. So at least any nonempty set carries a group structure. But abelianizing this particular group structure changes the cardinality.

Apologies if it is obvious, my group theory knowledge is just insufficient.

Possible Duplicate:
Does every non-empty set admit a group structure (in ZF)?

Let $X$ be an arbitrary nonempty set. Can you define a multiplication making it into an abelian group?

If $X$ is finite, say $|X|=n$, we can just use $X \cong \mathbb{Z}/n\mathbb{Z}$. What if $X$ is infinite?

If I'm not mistaken, the group of permutations of $X$ with finite support has the same cardinality as $X$. So at least any nonempty set carries a group structure. But abelianizing this particular group structure changes the cardinality.

Apologies if it is obvious, my group theory knowledge is just insufficient.

Possible Duplicate:
Does every non-empty set admit a group structure (in ZF)?

Let $X$ be an arbitrary nonempty set. Can you define a multiplication making it into an abelian group?

If $X$ is finite, say $|X|=n$, we can just use $X \cong \mathbb{Z}/n\mathbb{Z}$. What if $X$ is infinite?

If I'm not mistaken, the group of permutations of $X$ with finite support has the same cardinality as $X$. So at least any nonempty set carries a group structure. But abelianizing this particular group structure changes the cardinality.

Apologies if it is obvious, my group theory knowledge is just insufficient.

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Source Link
Community Bot
Community Bot
  • 1
  • 2
  • 3

Possible Duplicate:
Does every non-empty set admit a group structure (in ZF)?

Let $X$ be an arbitrary nonempty set. Can you define a multiplication making it into an abelian group?

If $X$ is finite, say $|X|=n$, we can just use $X \cong \mathbb{Z}/n\mathbb{Z}$. What if $X$ is infinite?

If I'm not mistaken, the group of permutations of $X$ with finite support has the same cardinality as $X$. So at least any nonempty set carries a group structure. But abelianizing this particular group structure changes the cardinality.

Apologies if it is obvious, my group theory knowledge is just insufficient.

Let $X$ be an arbitrary nonempty set. Can you define a multiplication making it into an abelian group?

If $X$ is finite, say $|X|=n$, we can just use $X \cong \mathbb{Z}/n\mathbb{Z}$. What if $X$ is infinite?

If I'm not mistaken, the group of permutations of $X$ with finite support has the same cardinality as $X$. So at least any nonempty set carries a group structure. But abelianizing this particular group structure changes the cardinality.

Apologies if it is obvious, my group theory knowledge is just insufficient.

Possible Duplicate:
Does every non-empty set admit a group structure (in ZF)?

Let $X$ be an arbitrary nonempty set. Can you define a multiplication making it into an abelian group?

If $X$ is finite, say $|X|=n$, we can just use $X \cong \mathbb{Z}/n\mathbb{Z}$. What if $X$ is infinite?

If I'm not mistaken, the group of permutations of $X$ with finite support has the same cardinality as $X$. So at least any nonempty set carries a group structure. But abelianizing this particular group structure changes the cardinality.

Apologies if it is obvious, my group theory knowledge is just insufficient.

Post Closed as "exact duplicate" by Steven Gubkin, Theo Johnson-Freyd, Andreas Blass, Emil Jeřábek, Ryan Budney
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Chris Heunen
Chris Heunen
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Can every nonempty set carry abelian group structure?

Let $X$ be an arbitrary nonempty set. Can you define a multiplication making it into an abelian group?

If $X$ is finite, say $|X|=n$, we can just use $X \cong \mathbb{Z}/n\mathbb{Z}$. What if $X$ is infinite?

If I'm not mistaken, the group of permutations of $X$ with finite support has the same cardinality as $X$. So at least any nonempty set carries a group structure. But abelianizing this particular group structure changes the cardinality.

Apologies if it is obvious, my group theory knowledge is just insufficient.