quantum field theory - Why do single particle states furnish a rep. of the inhomogeneous Lorentz group?

Following up on this question: Weinberg says

In general, it may be possible by using suitable linear combinations of the $\psi_{p,\sigma}$ to choose the $\sigma$ labels in such a way that $C_{\sigma'\sigma}(\Lambda, p)$ is block-diagonal; in other words, so that the $\psi_{p,\sigma}$ with $\sigma$ within any one block by themselves furnish a representation of the inhomogenous Lorentz group.

But why inhomogeneous Lorentz group if, in the first place, we performed a homogeneous Lorentz transformation on the states, via $U(\Lambda)$? I also want to be clear what is meant by the states "furnishing" a representation.

Regarding the above confusion, the same scenario again shows up during the discussion on the little group. Here's a little background: $k$ is a "standard" 4-momentum, so that we can express any arbitrary 4-momentum $p$ as $p^{\mu} = L^{\mu}_{\nu}(p) k^{\nu}$, where $L$ is a Lorentz transformation dependent on $p$. We consider the subgroup of Lorentz transformations $W$ that leave $k$ invariant (little group), and find that:

$U(W)\psi_{k \sigma} = \sum_{\sigma'} D_{\sigma' \sigma}(W)\psi_{k \sigma'}$. Then he says:

The coefficients $D(W)$ furnish a representation of the little group; i.e., for any elements $W$ and $W'$ , we get $D_{\sigma' \sigma}(W'W) = \sum_{\sigma''}D_{\sigma' \sigma''}(W)D_{\sigma''\sigma}(W')$.

So is it that even in the first part about the Lorentz group, $C$ matrices furnish the representation and not $\psi$?

Also, for the very simplified case if $C_{\sigma'\sigma}(\Lambda, p)$ is completely diagonal, would I be correct in saying the following in such a case, for any $\sigma$?

$$U(\Lambda)\psi_{p,\sigma} = k_{\sigma}(\Lambda, p)\psi_{\Lambda p, \sigma}$$

Only in this case it is clear to me that $U(\Lambda)$ forms a representation of Lorentz group, since $\psi_{p,\sigma}$ are mapped to $\psi_{\Lambda p, \sigma}$.