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.

**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.