Is the empty set a group?

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?

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

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.