Anindya - yes, the Eckmann-Hilton argument is a converse. I strongly urge you to prove it yourself, because it's delightful. I suggest proving this:

**Eckmann–Hilton Theorem.** If \$$M\$$ is a monoid with multiplication \$$\cdot : M \times M \to M\$$, and \$$\cdot : M \times M \to M\$$ is a homomorphism of monoids, then \$$M\$$ is commutative.

Here I'm making \$$M \times \$$ into a monoid in the obvious way, with multiplication

\$(a,b) \cdot (c,d) = (a \cdot c , b \cdot d).\$

You can prove it for groups if you feel more comfortable with them, but I recommend proving it for monoids instead of groups because inverses are not used in the argument. As far as I know, the identity element of the monoid _does_ play an essential role in the argument!

This is one of the most satisfying short proofs I know.