From now on $|i\rangle$ denotes the i-th element of the normalized canonical basis of the finite-dimensional Hilbert space $\mathbb{C}^n$ where the unitary nxn matrices $D(g)$ are defined. Furthermore, $$D_{ij}(g):= \langle i| D(g) |j\rangle.$$
Suppose that $D_{ij}(g) =0$ for all $g$ and a choice of $i,j$.
Notice that $i\neq j$ necessarily, otherwise we would have $|||i\rangle||=0 \neq 1$ when choosing $g=e$ (the unit element of the group).
The subspace $S$ of the finite linear combinations
$$\sum_{k=1}^N c_k D(g_k)|i\rangle\quad N=1,2,\ldots\:, \quad g_k \in H\:, c_k \in \mathbb{C}$$
is invariant under the action of $D$ trivially, because $$D(f)D(g_k)|j\rangle =D(g'_k)|i\rangle := D(f\circ g_k)|j\rangle$$ extended to the whole S by linearity. Regarding the space $S$,
(1) it does not coincide with the whole vector space since it is made of vectors orthogonal to $|i\rangle$;
(2) it is not made of the zero vector only, as it includes the vector $|j\rangle \neq 0$.
Hence $S$ is an invariant proper subspace of $\mathbb{C}^n$. In other words $D$ is reducible. In summary,
PROPOSITION.
If a unitary finite-dimensional representation $D$ on $\mathbb{C}^n$ ($n>1$) of a group $G$ is irreducible, then $D_{ij}(g) =0$ for all $g$ and a choice of $i,j$ cannot hold, where
$$D_{ij}(g):= \langle i| D(g) |j\rangle$$
and $|i\rangle$ is the generic element of the canonical basis of $\mathbb{C}^n$.
The thesis and the proof is still valid if the space is infinite dimensional the representation $D$ is unitary by referring to a fixed Hilbert basis. Just replace the subspace above with its closure.
What about the converse statement? What it is possible to prove is the following fact.
PROPOSITION. Let $G \ni g \mapsto D(g)$ be a representation of the group $G$ made of unitary operators $D(g): H \to H$ on the Hilbert space $H$ ($\dim(H)>1$, also infinite dimensional and non-separable). If $D$ is reducible, then there is a Hilbert basis $\{u_j\}_{j\in J} \subset H$ such that
$$\langle u_i| D(g) u_j\rangle =0 \quad \mbox{for all $g\in G$ and a pair $i,j \in J$ with $i\neq j$.}$$
PROOF. let $S \subset H$ be a (closed) proper invariant subspace, i.e. $D(g)(S) \subset S$ for every $g\in G$. Since the representation is unitary $D(g)(S^\perp) \subset S^\perp$ for every $g\in G$ and also $S^\perp$ is proper because $H= S\oplus S^\perp$ and $S \neq H$, $S\neq \{0\}$. Let $B$ be a Hilbert basis of $S$ and $B'$ a Hilbert basis of $S^\perp$. $B\cup B'$ is a Hilbert basis of $H$ by construction. Any pair of vectors one in $B$ and the other in $B'$ satisfies the thesis. QED
To every Lie algebra $\mathfrak{g}$ there is an associated simply-connected Lie group $G_s$. By the general correspondence of Lie groups and algebras, every representation of $G_s$ induces one of $\mathfrak{g}$ and vice versa, regardless of surjectivity of the exponential map. Hence a representation of $G_s$ is completely reducible if and only if the corresponding representation of $\mathfrak{g}$ is.
For any connected Lie group $G_c$ with Lie algebra $\mathfrak{g}$, this is still true: If we have a representation $V_r$ of $G_c$ that has a subrepresentation $W_r\subset V_r$ of $G_c$ and $\mathfrak{g}$ is semi-simple, then we get $V_r = W_r\oplus W'_r$ for some other representation $W'_r$ of $\mathfrak{g}$. But $W'_r$ must be a subrepresentation of $G_c$ also, since you can always use the exponential map locally to get the representation of $G_c$ from the representation of $\mathfrak{g}$ - very similar to the general proof of the algebra-group correspondence, surjectivity is not needed here either, only connectedness so that you can "shift" the exponential map along paths in the group.
For a disconnected Lie group $G_d$ with Lie algebra $\mathfrak{g}$, we have that all connected components are diffeomorphic to the identity component $G_0$ and there is an exact sequence
$$ 1 \to G_0 \to G \to \pi_0(G)=G/G_0\to 1. \tag{1}$$
If this sequence splits, we have $G = \pi_0(G)\rtimes G_0$ where $\rtimes$ is a semi-direct product. If $\pi_0(G)$ is additionally Abelian, Mackey's theory of induced representations tells us that every representation of $G$ comes from representations of the factors $\pi_0(G)$ and $G_0$ and every pair of irreducible representations of the factors defines an irreducible representation of $G$. Since Maschke's theorem tells us that finite group have only completely reducible finite-dimensional representations, this means that a Lie group for which (1) splits and for which $\pi_0(G)$ is Abelian has completely reducible representations if $\mathfrak{g}$ is semi-simple.
Applying this to the Lorentz group, we have $\pi_0(G) = \mathbb{Z}_2\times\mathbb{Z}_2$ as the group of space parity and time reversal. It is Abelian, and one can show that the Lorentz group is indeed the semi-direct product of this and its connected component by explicitly writing out its matrix structure. Therefore, all finite-dimensional representations of the Lorentz group are completely reducible, and its irreducible representations can be given by an irreducible representation of the proper orthochronous Lorentz group $(V,\rho)$ + an irreducible representation of time reversal (which under the tensor product of the induced representation becomes just some operator on $V$ that squares to the identity) + an irreducible representation of space parity (likewise).
Best Answer
Every representation $(D,V)$ of a finite group $G$ is equivalent to a unitary representation.
It is often termed as Weyl's unitary trick. This works by simply redefining your inner product by averaging over on the space $V$. This smoothening trick works precisely because of finite number of elements and invariance of sum of finite elements (i.e. $\sum_{g \in G} D(gh) = \sum_{gh \in G} D(gh) = \sum_{g \in G} D(g)$).
$$\langle v| w \rangle = \frac{1}{|G|}\sum_{g \in G} \langle D(g)v| D(g)w \rangle$$
You can verify how $\langle v| w \rangle = \langle D(h)v| D(h)w \rangle$ for some $h \in G$. In fact, this claim is similar to that of existence of a common basis change matrix $S$ such that $D'(g) = SD(g)S^{-1}$ is a unitary representation.