Eckmann-Hilton argument

Proposition. (Eckmann-Hilton). Let $G$$G$ be a set and $\ast ,\circ$$*,\circ$ be two untial operator with identity $1_{\ast },1_{\circ }$$1_*,1_{\circ}$. And for any $a,b,c,d\in G$$a,b,c,d\in G$,

$$\begin{array}{}\text{(1)}& (a\circ b)\ast (c\circ d)=(a\ast c)\circ (b\ast d).\end{array}$$

Then $1_{\ast }=1_{\circ }$$1_*=1_\circ$, $\ast =\circ$$*=\circ$ and it is commutative and associative.

Proof.

$$\begin{array}{}\text{(2)}& 1_{\ast }=1_{\ast }\ast 1_{\ast }=(1_{\ast }\circ 1_{\circ })\ast (1_{\circ }\circ 1_{\ast })=(1_{\ast }\ast 1_{\circ })\circ (1_{\circ }\ast 1_{\ast })=1_{\circ }\circ 1_{\circ }=1_{\circ }\end{array}$$

So there exists only one identity, we denote it as $1$$1$.

$$\begin{array}{}\text{(3)}& a\ast b=(a\circ 1)\ast (1\circ b)=(a\ast 1)\circ (1\ast d)=a\circ d\end{array}$$

Hence $\ast =\circ$$*=\circ$.

Then

$$\begin{array}{}\text{(4)}& (1a)(b1)=(b1)(a1)=ba\end{array}$$

Finally,

$$\begin{array}{}\text{(5)}& (ab)c=(ab)(1c)=(a1)(bc)=a(bc)\end{array}$$

Corollary. Let $(G,\cdot )$$(G,\cdot)$ be a topological group, then ${\pi }_1(G)$$\pi_1(G)$ is abelian group.

Proof. Let $\circ$$\circ$ be the multiplication of ${\pi }_1(G)$$\pi_1(G)$ and $\cdot$$\cdot$ be the multiplication of path pointwise, i.e. $(\alpha \cdot \beta )(t)=\alpha (t)\cdot \beta (t)$$(\alpha\cdot\beta)(t)=\alpha(t)\cdot\beta(t)$.

Then

$$\begin{array}{}\text{(6)}& (\alpha \cdot \beta )\circ ({\alpha }^{\prime }\cdot {\beta }^{\prime })=(\alpha \circ {\alpha }^{\prime })\cdot (\beta \circ {\beta }^{\prime })\end{array}$$

Hence $\cdot =\circ$$\cdot=\circ$ and it is commutative! $◻$$\square$