To be really fair, a group is a lot like the product of a monoid with itself as an opposite category.

Especially since for any \\(f\\) in our monoid \\(\mathbf{M}\\),

\\[f^{op} \circ f = id\_{\mathbf{M}} = f\circ f^{op}.\\]

Especially since for any \\(f\\) in our monoid \\(\mathbf{M}\\),

\\[f^{op} \circ f = id\_{\mathbf{M}} = f\circ f^{op}.\\]