= Solution
The identity is one <conjugacy class>. In $N\setminus\{1\}$, conjugation by $c$ pairs each element with its inverse, giving four classes represented by $a,b,ab,ab^2$, each of size two. All nine elements of $Nc$ are conjugate, since conjugation by $n\in N$ changes the reflection component by $n^2$ and squaring is a bijection of $N$.
There are two <linear characters>: the trivial character and the character equal to one on $N$ and minus one on $Nc$. Write $\omega=e^{2\pi i/3}$. The remaining four irreducible <characters of a representation> are induced from the nontrivial characters $a^i b^j\mapsto\omega^{ri+sj}$ of $N$, pairing $(r,s)$ with $(-r,-s)$. They are exactly the two-dimensional representations already proved irreducible. Taking representatives $(1,0),(0,1),(1,1),(1,2)$ yields the <character table>
$$
\begin{array}{c|rrrrrr}
&1&a&b&ab&ab^2&c\\\hline
\text{class size}&1&2&2&2&2&9\\\hline
1&1&1&1&1&1&1\\
\mathrm{sgn}&1&1&1&1&1&-1\\
\chi_{10}&2&-1&2&-1&-1&0\\
\chi_{01}&2&2&-1&-1&-1&0\\
\chi_{11}&2&-1&-1&-1&2&0\\
\chi_{12}&2&-1&-1&2&-1&0
\end{array}
$$
Indeed the induced value on $a^i b^j$ is $\omega^{ri+sj}+\omega^{-(ri+sj)}$, equal to two or minus one, and is zero off $N$. The squared dimensions sum to $1+1+4\cdot4=18$, so this is the complete table.
An element of $N$ centralizes $c$ only if it equals its inverse, hence only if it is the identity. No reflection centralizes $a$. Thus $Z(E)=\{1\}$ is cyclic, but the two linear representations kill $N$ and each two-dimensional irreducible has a nontrivial <kernel of a group homomorphism> of order three. Therefore \b[a cyclic centre does not imply a faithful <irreducible representation>].
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