The idea that different particles 'are' irreducible representations of a group, or that they 'are' the group elements themselves, is a shorthand for a more complicated situation. Particles are excitations of a field that obeys a differential equation; something like the Schrodinger equation. If the differential equation has certain symmetries, expressed by a symmetry group, then its solutions must in some sense have the same symmetries. We can deduce a lot about the solutions and how they are interrelated by studying the symmetry group and its representations, to the point where we never even bother to work out the differential equations and its solutions explicitly - we only look at the group and its properties.
I'm going to argue by analogy with the situation we have for the Schrodinger equation of a Hydrogen atom. I think the truth is probably more complicated than this, but this gives the gist.
So suppose we have a differential equation of the form $H\psi=\epsilon \psi$, where $H$ is a differential operator, $\epsilon$ is a constant number, and $\psi$ is the wavefunction. The possible solutions for $\epsilon$ are the eigenvalues, and for each eigenvalue there are one or several possible solutions $\psi$ which we call eigenvectors or eigenfunctions or eigenstates. For the Schrodinger equation, $H$ is the Hamiltonian, $\epsilon$ is the energy, and the wavefunction evolves in time according to the differential equation $H\psi=i\hbar \partial \psi / \partial t$.
The Hamiltonian for a particle in a potential well might look something like:
$$H(\mathbf{r})=-\frac{\hbar^2}{2\mu}\left( \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2} \right)-\frac{e^2}{\sqrt{x^2+y^2+z^2}}$$
The first term on the right is the kinetic energy of the particle, the second term is the spherically symmetric potential well of the Hydrogen atom, in which an electron sits.
This potential is spherically symmetric. If we rotate the position vector about the origin, the Hamiltonian doesn't change. $H(T\mathbf{r})=H(\mathbf{r})$. The set of transformations $T \in \mathcal{G}$ for which the Hamiltonian doesn't change form a group, called the invariance group of the Hamiltonian. In this case, it is the 3D rotation group, but it doesn't have to be. Other potential well shapes will give rise to other symmetry groups.
We can apply the same transformations to the wavefunction that we just applied to the Hamiltonian operator. We write $(P(T)\psi)(\mathbf{r})=\psi(T\mathbf{r})$, where $P(T)$ is the operator that applies the coordinate transformation $T$ to the wavefunction. The operators $P(T)$ for $T \in \mathcal{G}$ form a group isomorphic to $\mathcal{G}$. We use a slightly different notation for them as they're a different mathematical object, acting on wavefunctions rather than coordinates, but it's the same set of transformations.
If we have a set of $d$ linearly-independent wavefunctions $\psi_1(\mathbf{r}),\psi_2(\mathbf{r}),\ldots,\psi_d(\mathbf{r})$, then they can be treated as the basis vectors of a $d$-dimensional representation $\Gamma$ of $\mathcal{G}$ if for every coordinate transformation $T$ of $\mathcal{G}$,
$$P(T)\psi_n(\mathbf{r})=\sum_{m=1}^d {\Gamma(T)_{mn}\psi_m(\mathbf{r})}$$
This is saying that when we transform one of our basis vectors, we get some linear combination of it and the other basis vectors as a result. The coefficients $\Gamma_{mn}$ of the linear combination for each basis vector can be arranged in a matrix, and multiplying the matrices performs the same action on any linear combination of basis vectors as applying the coordinate transformation.
And finally, going back to our eigenvalue equation $H(\mathbf{r})\psi(\mathbf{r})=\epsilon\psi(\mathbf{r})$, if we have a complete set of $d$ eigenfunctions $\psi_i(\mathbf{r})$ that all correspond to the same eigenvalue $\epsilon$, then they form a basis for a $d$-dimensional representation of the invariance group of the equation. Transforming any one of them in a way that leaves the Hamiltonian invariant gives another eigenfunction with the same eigenvalue, which is therefore in the same vector space, and which therefore has to be a linear combination of the basis eigenfunctions.
This was the reasoning that was used to help solve the Schrodinger equation for a Hydrogen atom using its spherical symmetry as a shortcut - the orbital solutions use the spherical harmonic functions as a basis. Each energy level has a different number of orbitals. The $d$ orbitals in an energy level are basis vectors of a $d$-dimensional representation of the rotation group. In the same way, we identify the internal symmetry group of the field that all particles are excitations of, which comes from the symmetries of whatever differential equation it obeys, and then the irreducible representations tell us about how the solution wavefunctions might be related. A $d$-dimensional representation corresponds to $d$ solution eigenfunctions all corresponding to the same eigenvalue.
The mental picture this ought to evoke would be to consider the group as like the way we know the Hydrogen atom is spherically symmetric, so rotations leave it unchanged, and the different particles are more like the different electron shells and orbitals in a Hydrogen atom. Each 'particle' is the universal wavefunction wobbling in a different pattern. But we don't look directly at the individual patterns, only at how they are related to the overall symmetry.
