= Solution
The <Klein four-group> has multiplication table
$$
\begin{array}{c|cccc}
&1&a&b&c\\\hline
1&1&a&b&c\\
a&a&1&c&b\\
b&b&c&1&a\\
c&c&b&a&1
\end{array}.
$$
Since it is abelian, every <conjugacy class> is a singleton: $\{1\},\{a\},\{b\},\{c\}$. An <irreducible representation> is a nonzero representation whose only invariant subspaces are zero and the whole representation space. For a <finite group>, the number of inequivalent complex irreducible representations equals the number of conjugacy classes, so there are four here.
A one-dimensional representation sends the involutions $a,b$ to signs, independently. All four choices are
$$
\chi_{pq}(1)=1,\quad\chi_{pq}(a)=(-1)^p,\quad\chi_{pq}(b)=(-1)^q,\quad\chi_{pq}(c)=(-1)^{p+q},\qquad p,q\in\{0,1\}.
$$
They are distinct irreducible representations, exhausting the four possibilities. The <character table of the Klein four-group> is
$$
\boxed{\begin{array}{c|rrrr}
&1&a&b&c\\\hline
\chi_{00}&1&1&1&1\\
\chi_{01}&1&1&-1&-1\\
\chi_{10}&1&-1&1&-1\\
\chi_{11}&1&-1&-1&1
\end{array}.}
$$
The specified diagonal <matrices> obey $D(a)^2=D(b)^2=I$, commute, and satisfy $D(a)D(b)=D(c)$, so the group relations prove that $D$ is a <group representation>. Its coordinate axes are invariant lines carrying respectively $\chi_{01},\chi_{10},\chi_{11}$. Therefore
$$
\boxed{D\cong\chi_{01}\oplus\chi_{10}\oplus\chi_{11}.}
$$
The <trivial representation> $\chi_{00}$ is absent. If $H$ denotes the four-row character <matrix> above, direct row products give $HH^T=4I$. This verifies <character orthogonality>:
$$
\frac14\sum_{g\in V}\overline{\chi_{pq}(g)}\chi_{rs}(g)=\delta_{pr}\delta_{qs}.
$$
The character of $D$ is $(3,-1,-1,-1)$, whose inner products with the four rows are $0,1,1,1$, independently confirming the decomposition. All these irreducibles are real as well: commuting real involutions can be simultaneously diagonalized into their real sign eigenspaces.
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