To form the group according to the requirements. From VGT by Nathan Carter 
Now for c) i think of a set with permutations on $a,b,c$. The three actions can be $f_1$= swapping first and second number. $f_2$= swapping second and third number and $f_3$ means swap of third and first number. But i am not sure
 A: Here are some examples that fit the criteria.  Hopefully you'll be able to justify to yourself that these indeed work.


*

*(a): Take your actions to be all possible permutations on the letters $a,b,c$.

*(b): Take your actions to be all possible permutations on the letters $a,b$ (so either we swap or we do nothing)

*(c): Our object will be an equilateral triangle with one corner pointing upwards.  Take your actions to be the rotations that result in the triangle still pointing upwards.

*(d): The same as (c), but with a square.

*(e): Consider all possible rotations of a circle. Another example is all possible translations of an object along the number-line.


Your example for (c) does not work since it is not qualify as a group.  Note that (following rule 1.5) your list of actions is $\{f_1,f_2,f_3\}$.  With this setup, we don't satisfy rule 1.8, that every sequence of consecutive actions is also an action (in more conventional terminology, we would say that your set fails to be closed under composition, which is your set's binary operation).  In particular, consider the effect of applying $f_1$ then $f_2$.
$$
(a,b,c) \overset{f_1}\to (b,a,c) \overset{f_2}\to (b,c,a).
$$
The cyclic permutation $(a,b,c) \to (b,c,a)$ is not on our list of actions.
Note that if we were to make a group by filling in your list with all possible outcomes (i.e. if we were to consider the group "generated by" the elements $f_1,f_2,f_3$), then we would end up with the answer that I gave for (a).
