Hmm but in a field you can't divide by zero, so you are demanded a multiplicative group only after you strip zero away. In \\((\mathbb{Z}/2, \cdot)\\) you have two actions in \\(\\{0,1\\}\\): the identity, and the one that zeroes all (and obeys \\(f \circ f = f\\)). You exclude the later, and you get a singleton group where the identity is its own inverse. You have operation tables [here](https://en.wikipedia.org/wiki/GF(2)).

EDIT: to actions -> *two* actions

EDIT: to actions -> *two* actions