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I was wondering if an empty set can be a torsion group (since the definition of torsion group is that if $x$ is in the set $X$ has a finite order. However, the assumption is false, so the implication is true) and with same reasoning, torsion-free?

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    $\begingroup$ The empty set is not a group at all, as it does not contain a neutral element. $\endgroup$ Nov 26, 2014 at 15:38

4 Answers 4

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You probably are confusing the empty set (which is not a group) with the trivial group, having only the identity as element. This group is indeed torsion (any element has finite order) and torsionfree (any non identity element has infinite order).

In this way the statements

every subgroup of a torsionfree group are torsionfree

and

every quotient of a torsion group is torsion

are valid without restrictions.

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    $\begingroup$ It worth mentioning that, although for the trivial group it is (formally) correct that ∀ x: (x ≠ e) → x isn’t a torsion, there are no such x that x ≠ e . $\endgroup$ Dec 12, 2014 at 9:50
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The empty set cannot be a torsion group nor can it be a torsion-free group because the empty set cannot be a group. A group must have an identity element.

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Hints:

Does an emptyset has an identity element?

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The empty set is not a group because there is no identity element. I think maybe you are confused about the definition of a group. Many people get confused the idea that there must exists an identity element in the set, vs for all $g\in G \ \exists ! e \in G$ s.t. $g*e=e*g=g$. If we used this second statement for the definition of a group, then the empty set would be a group(by vacuous truth).

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  • $\begingroup$ Identity element is usually understood as an attribute of a structure (constant ≃ 0-ary operation), not as an axiom. Certainly, empty set can’t be a model for any mathematical structure with constants. $\endgroup$ Dec 12, 2014 at 9:53

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