Singleton Groups?

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Let G = {0}. Let + be the usual addition such that a+b=c.

(1) The operation + is well-defined.
Suppose a₁+b₁=c₁ and a₂+b₂=c₂. Then a₁=a₂, b₁=b₂, and c₁=c₂. So c₁=a₁+b₁=a₂+b₂=c₂.

(2) The operation + is closed.
Pick a, b ∈ G. Then a+b = 0+0 = 0 = c. Thus c ∈ G, and + is closed in G.

(3) The operation is associative.
Pick a, b, c ∈ G. Then (a+b)+c = (0+0)+0 = 0+(0+0) = a+(b+c).

(4) There is an identity element.
Pick a ∈ G. Then a+0 = a, for all 0 ∈ G.

(5) Every element of G has an inverse under +.
Pick a ∈ G. Then (-a) ∈ G and a+(-a) = 0. Thus (-a) functions as the additive inverse of G.

Let H = {1}. Let ◦ denote the usual multiplication of real numbers on H such that a ◦ b = c.

(1) The operation ◦ is well defined.
Suppose a₁◦ b₁ = c₁ and a₂◦ b₂ = c₂. Then a₁=a₂, b₁=b₂, and c₁=c₂. So c₁=a₁◦ b₁=a₂◦ b₂=c₂.

(2) The operation ◦ is closed.
Pick a, b ∈ H. Then a ◦ b = 1 ◦ 1 = 1. Since 1 ∈ H, ◦ is closed in H.

(3) The operation ◦ is associative.
Pick a, b, c ∈ H. Then (a ◦ b) ◦ c = (1 ◦ 1) ◦ 1 = 1 ◦ (1 ◦ 1) = a ◦ (b ◦ c).

(4) There is an identity element.
Pick a ∈ H. Then a ◦ 1 = a. Thus 1 functions as the identity element.

(5) Every element of H has an inverse under ◦.
Pick a ∈ H. Then a ◦ (1/a) = 1. Thus (1/a) = a⁻¹ ∈ H, and every element has an inverse in H.


And it is trivial to show that both G and H are Abelian groups.

I've worked this out on my own, following a thought I had about groups of a single element. Are these groups even considered in Group theory, or are they just tossed aside as useless knowledge. If they are considered in Group theory, what other purposes do they serve?

take your word for it, all I recognize is the Klein bottle...

There is only one trivial group. The two you defined are isomorphic, which means that they are the same up to relabeling stuff. The trivial group is not that interesting or useful, but yes, group theorists are aware that it exists.

Nice Klein bottle, though.