Given groups $H, K$ and $\cdot : H × K \to K$ a group action, is $a \cdot b × a \cdot c = a \cdot (b × c)$?
Here a group action is $\cdot : H × K \to K$ such that for all suitable elements $h \cdot (h' \cdot k) = hh' \cdot k$ and $1 \cdot k = k$.
Here, × is the group operator of $K$. This at first seemed quite obvious, but trying to prove it proved futile. Is this even true? Certainly this is true for the conjugation action but what about a general action? If not a counterexample would be great.
If $H$ acts on $K$ as a set, then this is false in general. For example, let $|H|=|K|=2$ with $H$ acting on $K$ transitively, and $a$ the non-identity element of $K$, and $b$ the non-identity element of $H$. Then $b\cdot a ~\times ~b\cdot 1 = 1\times a = a$, which is not equal to $b\cdot(a\times 1)=b\cdot a=1$.
In fact, when your statement does hold, we call $K$ a group with operators (the operators being from $H$, which need not be a group for purposes of this paragraph). There are analogues of many theorems for ordinary groups...e.g., a finite group with operators has a composition series of subnormal subgroups (respectively a chief series of normal subgroups) that are preserved by the operators (and which may have fewer terms than the ordinary composition series and chief series, respectively).
If, in addition to the above paragraph, $K$ is abelian and $H$ is again a group, we call $K$ an $H$-module.