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I know that a set G with a binary operation $*$ is a group, if:

  1. $a*b\in G$, for all $a, b \in G$.

  2. $*$ is associative:

$$(a*b)*c=a*(b*c) \\ \text{for all }a, b, c\in G.$$

  1. An identity element $e \in G$ exists, such that

$$a*e = e*a = a\\ \text{for all }a\in G.$$

  1. For all elements $a \in G$, there exists an $a^{-1} \in G$, such that:

$$a*a^{-1} = a^{-1}*a=e.$$

Can I use that to show that the empty set is a group?

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    $\begingroup$ Hint: There is only one possible binary operation on the empty set. Can that operation have an identity element? $\endgroup$
    – qaphla
    May 19, 2016 at 1:54
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    $\begingroup$ No. A group contains the identity element. $\endgroup$
    – BrianO
    May 19, 2016 at 1:54
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    $\begingroup$ It's a semigroup. $\endgroup$ May 19, 2016 at 4:15

2 Answers 2

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So we know that universal statements are true on empty domains, and existence statements are false. Being a group requires the existence of an identity element, and since the empty set cannot satisfy this (it has no elements) it is not a group.

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As BrianO says, $\emptyset$ is not a group, because every group has an identity element. This also means that $\emptyset$ is not a vector space, it's not a ring, it's not a module, and it's not a boolean algebra. However, $\emptyset$ is a perfectly good: semilattice, band, and affine space. Also, it's best to drop the non-emptiness condition from the usual definition of a heap, in which case $\emptyset$ is a perfectly good heap.

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