Eckmann-Hilton Argument

The Eckmann-Hilton Argument states that for any two Binary Operations ⋄ , ⋆ : A × A → A \diamond,\star:A\times A\to A with units 1 ⋄ , 1 ⋆ : A : subscript 1 ⋄ subscript 1 ⋆ 𝐴 1_{\diamond},1_{\star}:A such that ⋆ ⋆ \star is a homomorphism with respect to ⋄ ⋄ \diamond , then

The proof is an exercise in some easy algebra:

First, note that ⋆ : A × A → A \star:A\times A\to A the statement that is homomorphism means that it must be a homomorphism from the Product Magma A × A 𝐴 𝐴 A\times A to A 𝐴 A , and thus

If we unfold this, we get the following interchange law

Arumed with this observation, let us show that 1 ⋄ = 1 ⋆ subscript 1 ⋄ subscript 1 ⋆ 1_{\diamond}=1_{\star}

1⋄subscript1⋄\displaystyle 1_{\diamond} =1⋄⋄1⋄absent⋄subscript1⋄subscript1⋄\displaystyle=1_{\diamond}\diamond 1_{\diamond}
=(1⋄⋆1⋆)⋄(1⋆⋆1⋄)absent⋄⋆subscript1⋄subscript1⋆⋆subscript1⋆subscript1⋄\displaystyle=(1_{\diamond}\star 1_{\star})\diamond(1_{\star}\star 1_{\diamond})
=(1⋄⋄1⋆)⋆(1⋆⋄1⋄)absent⋆⋄subscript1⋄subscript1⋆⋄subscript1⋆subscript1⋄\displaystyle=(1_{\diamond}\diamond 1_{\star})\star(1_{\star}\diamond 1_{% \diamond})
=1⋆⋆1⋆absent⋆subscript1⋆subscript1⋆\displaystyle=1_{\star}\star 1_{\star}
=1⋆absentsubscript1⋆\displaystyle=1_{\star}

Next, x ⋄ y = x ⋆ y ⋄ 𝑥 𝑦 ⋆ 𝑥 𝑦 x\diamond y=x\star y .

x⋄y⋄𝑥𝑦\displaystyle x\diamond y =(x⋆1⋆)⋄(1⋆⋆y)absent⋄⋆𝑥subscript1⋆⋆subscript1⋆𝑦\displaystyle=(x\star 1_{\star})\diamond(1_{\star}\star y)
=(x⋄1⋆)⋆(1⋆⋄y)absent⋆⋄𝑥subscript1⋆⋄subscript1⋆𝑦\displaystyle=(x\diamond 1_{\star})\star(1_{\star}\diamond y)
=(x⋄1⋄)⋆(1⋄⋄y)absent⋆⋄𝑥subscript1⋄⋄subscript1⋄𝑦\displaystyle=(x\diamond 1_{\diamond})\star(1_{\diamond}\diamond y)
=x⋆yabsent⋆𝑥𝑦\displaystyle=x\star y

On to commutativity.

x⋆y⋆𝑥𝑦\displaystyle x\star y =(1⋄⋄x)⋆(y⋄1⋄)absent⋆⋄subscript1⋄𝑥⋄𝑦subscript1⋄\displaystyle=(1_{\diamond}\diamond x)\star(y\diamond 1_{\diamond})
=(1⋄⋆y)⋄(x⋆1⋄)absent⋄⋆subscript1⋄𝑦⋆𝑥subscript1⋄\displaystyle=(1_{\diamond}\star y)\diamond(x\star 1_{\diamond})
=(1⋆⋆y)⋄(x⋆1⋆)absent⋄⋆subscript1⋆𝑦⋆𝑥subscript1⋆\displaystyle=(1_{\star}\star y)\diamond(x\star 1_{\star})
=y⋄xabsent⋄𝑦𝑥\displaystyle=y\diamond x

Finally, associativity.

x⋆(y⋆z)⋆𝑥⋆𝑦𝑧\displaystyle x\star(y\star z) =(x⋄1⋄)⋆(y⋆z)absent⋆⋄𝑥subscript1⋄⋆𝑦𝑧\displaystyle=(x\diamond 1_{\diamond})\star(y\star z)
=(x⋄1⋄)⋆(y⋄z)absent⋆⋄𝑥subscript1⋄⋄𝑦𝑧\displaystyle=(x\diamond 1_{\diamond})\star(y\diamond z)
=(x⋆y)⋄(1⋄⋆z)absent⋄⋆𝑥𝑦⋆subscript1⋄𝑧\displaystyle=(x\star y)\diamond(1_{\diamond}\star z)
=(x⋆y)⋄(1⋆⋆z)absent⋄⋆𝑥𝑦⋆subscript1⋆𝑧\displaystyle=(x\star y)\diamond(1_{\star}\star z)
=(x⋆y)⋄zabsent⋄⋆𝑥𝑦𝑧\displaystyle=(x\star y)\diamond z
=(x⋆y)⋆zabsent⋆⋆𝑥𝑦𝑧\displaystyle=(x\star y)\star z

Corollaries