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}

1subscript1\displaystyle 1_{\diamond} =11absentsubscript1subscript1\displaystyle=1_{\diamond}\diamond 1_{\diamond}
=(11)(11)absentsubscript1subscript1subscript1subscript1\displaystyle=(1_{\diamond}\star 1_{\star})\diamond(1_{\star}\star 1_{\diamond})
=(11)(11)absentsubscript1subscript1subscript1subscript1\displaystyle=(1_{\diamond}\diamond 1_{\star})\star(1_{\star}\diamond 1_{% \diamond})
=11absentsubscript1subscript1\displaystyle=1_{\star}\star 1_{\star}
=1absentsubscript1\displaystyle=1_{\star}

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

xy𝑥𝑦\displaystyle x\diamond y =(x1)(1y)absent𝑥subscript1subscript1𝑦\displaystyle=(x\star 1_{\star})\diamond(1_{\star}\star y)
=(x1)(1y)absent𝑥subscript1subscript1𝑦\displaystyle=(x\diamond 1_{\star})\star(1_{\star}\diamond y)
=(x1)(1y)absent𝑥subscript1subscript1𝑦\displaystyle=(x\diamond 1_{\diamond})\star(1_{\diamond}\diamond y)
=xyabsent𝑥𝑦\displaystyle=x\star y

On to commutativity.

xy𝑥𝑦\displaystyle x\star y =(1x)(y1)absentsubscript1𝑥𝑦subscript1\displaystyle=(1_{\diamond}\diamond x)\star(y\diamond 1_{\diamond})
=(1y)(x1)absentsubscript1𝑦𝑥subscript1\displaystyle=(1_{\diamond}\star y)\diamond(x\star 1_{\diamond})
=(1y)(x1)absentsubscript1𝑦𝑥subscript1\displaystyle=(1_{\star}\star y)\diamond(x\star 1_{\star})
=yxabsent𝑦𝑥\displaystyle=y\diamond x

Finally, associativity.

x(yz)𝑥𝑦𝑧\displaystyle x\star(y\star z) =(x1)(yz)absent𝑥subscript1𝑦𝑧\displaystyle=(x\diamond 1_{\diamond})\star(y\star z)
=(x1)(yz)absent𝑥subscript1𝑦𝑧\displaystyle=(x\diamond 1_{\diamond})\star(y\diamond z)
=(xy)(1z)absent𝑥𝑦subscript1𝑧\displaystyle=(x\star y)\diamond(1_{\diamond}\star z)
=(xy)(1z)absent𝑥𝑦subscript1𝑧\displaystyle=(x\star y)\diamond(1_{\star}\star z)
=(xy)zabsent𝑥𝑦𝑧\displaystyle=(x\star y)\diamond z
=(xy)zabsent𝑥𝑦𝑧\displaystyle=(x\star y)\star z

Corollaries