Characteristics of a Character table and what it tells me. I am trying to solve the character table and some related questions. The questions are below, and what I have done is below that. Any help on any pieces I am sure will enlightening.

For parts c and d, I could really use some hints as I don't know where to begin.
e. The best I have figured out for the character table is
There are 5 conjugacy classes hence 5 irreducible representations. There 20 elements in the group and the only way to break this down to the sum of squares is $1^2 + 1^2 + 1^2 + 1^2 + 4^2$ so these are the orders of the conjugacy classes of the trivial element.
\begin{array}{rrrrrrrrrrr}
 & C_1 & C_2 & C_3 & C_4  & C_5\\
\chi_0 & 1 & 1 & 1 & 1 & 1\\
\chi_1 & 1&  1& i & -i & -1 &   \\
\chi_2 & 1 & 1 & -i &  -i &  -1&   \\
\chi_2 & 1 &1 &-1 &  -1& 1 &   \\
\chi_3 &  4 & -1 &  0&  0&  0&   \\
C_G(x) & 1 & 5 & 4 & 4 & 4 &
  \end{array}
although I think $d,e$ should be a -i and i pair. With my calculations I am having a hard time just putting in numbers trying to satisfy $\left \langle \alpha, \beta\right \rangle := \frac{1}{|G|}\sum_{g \in G} \alpha(g) \overline{\beta(g)} $ and the column version of Schur's orthogonality relationship>
f. I see that the centralizer of $C_3$ has 4 elements but does this, and if so how, does this mean that elements $x \in C_3$ have order divisible by 4? 
g. more properties of conjugacy classes and that somehow I can read this from the character table?
Really thanks for any hints, tips, or answers.
 A: I am answering this question using Sylow's theorem as a crutch - I don't know whether that is 'allowed.' Still, I hope this helps. 
Sylow says that the number 5-Sylow groups is a factor of $20/5$, and is congruent to $1 \pmod 5$. Therefore, there is a normal subgroup $S$ of order $5$, necessarily cyclic, in $G$, $S = C_1 \cup C_2$, and we have the exact sequence 
$$ 1 \rightarrow S \rightarrow G \rightarrow H \rightarrow 1,$$
where $H$ is a group of order $4$ and thus abelian. Therefore, there are $4$ 1-dimensional representations of $G$ which factor through $H$, i.e., the $4$ irreducible representations of $H$; since the table (completed by you) tells us that $i$ is one of the character values, we know that $H \simeq \mathbb Z /4Z$, and that gives four (multiplicative) characters: 1, $\chi$, $\chi^2$, and $\chi^3$, with $\chi(1) =\chi(C_2)=1$, $\chi(C_3) =i$, $\chi(C_4) =-i$, and $\chi(C_5) = -1$.
By Sylow again, the above sequence splits, i.e., $G \simeq S \rtimes H $. From the conjugacy class counts, in particular the cardinality of $C_2$, we know that $H$ maps onto ${\rm Aut}(S)$.
At this point,  we know what $G$ is up to isomorphism: $G$ is isomorphic to the group of affine transformations $g_{{a,b}}\colon x \mapsto ax +b$, $a\in \def \F {{\mathbb F_5}} \F^*$ and $b\in \F$:  $C_1 =1$, $g_{(1,1)}\in C_2$, $g_{(2,0)}\in C_3$, $g_{(3,0)}\in C_4$, and $g_{(4,0)}\in C_5$. (This deals with f and g of the question.) The commutator group $[G,G] = S$: on the one hand, $H$ is abelian, so $G^{ab}\rightarrow H$, and the commutator subgroup is a subgroup of $S$; on the other hand, $G$ is not abelian.     
You already wrote down the 'missing' character (using orthogonality and arithmetic), but as a (less good) alternative:  As $S\simeq \mathbb Z/5$, we know the 5 (multiplicative) characters of $S$, $\psi^k$, $k\in \mathbb Z/5$.  The induced character 
$$ \Psi(g) = \sum_{k=1}^4 \psi' \left( g_{(k,0)}^{-1}\, g\, g_{(k,0)}\right),$$
where $\psi' (s) = \psi ( s)$ if $s\in S$ and $0$ otherwise, is the missing character: $\Psi( 1) = 4$, $\Psi ( C_2) = -1$ (because the sum of the non-trivial fifth roots of unity = $-1$ ), and zero otherwise.   
