decomposition of representation kG of G 
Decompose $kG$ in to indecomposable representations and decide which summands are irreducible.
(a)$G=S_2,k=\mathbb{C}$
(b)$G=\mathbb{Z}/3\mathbb{Z},k=\mathbb{C}$
(c)$G=\mathbb{Z}/3\mathbb{Z},k=\mathbb{F}_3$

ideas: Maschke's Theorem tells us that (a) and (b) are completely decomposable, but (c) is not. For (a): If $S_2=\{e,s\}$, then $span(e),span(s)$ should be the subrepresentations.
For (b) and (c) I have no idea, how does $\mathbb{Z}/3\mathbb{Z}$ act?
 A: As was already answered in the comments, the action is always left multiplication (or right multiplication, depending on your preference of left or right modules).
Some hints:
(a): $\operatorname{span}(e)$ is not a subrepresentation of $kG$ in this case since $s\cdot e=s\notin \operatorname{span}(e)$. But you are right, this two-dimensional representation is the direct sum of two one-dimensional representations. To find out which ones, take an arbitrary element of $kG$, i.e. $\alpha e+\beta s$ for $\alpha,\beta\in \mathbb{C}$. Now multiply with $s$, then you get $\beta e+\alpha s$, but this should be in the span of $\alpha e+\beta s$. So the question is, what can you choose $\alpha$ and $\beta$ to be in order to satisfy that.
(b): The idea is the same as in (a), only now you have a three-dimensional representation with basis $e, s, s^2$. Here you also need to use complex numbers, whereas in (a) $\mathbb{Q}$ would suffice. 
(c): That's a bit trickier. I'll give to hints: 1) This representation is indecomposable. 2) Consider powers of the element $(1-g)$, where $g$ is a generator of the group.
