What does "action of a gauge group on a particle" mean? I have come across this phrase

left handed fermions transform under $SU(3)\times{}SU(2)\times{}U(1)$ differently from the way right handed fermions do.

I am just beginning to learn about how the particles of the standard model are related to irreps of the gauge group. So, I would like to know what the phrase transforming under a gauge group exactly means, and what's the meaning of an action of a gauge group on whatever it acts on.
 A: Basic Lie Facts
A group is a general collection of abstract entities; in physics we concern ourselves with those  of unitary operators which act on vector spaces of quantum states. Continuously generated groups, for which any element is arbitrarily close to the identity, by the repeated application of infinitesimal elements, are denoted Lie groups. Any infinitesimal element may be written as,
$$g=1+i\alpha_a T^a + \dots$$
where $T^a$ are hermitian generators of the group $G$. These must satisfy a relation,
$$[T^a,T^b]=if^{ab}_cT^c$$
where $f^{ab}_c$ are the structure constants of the group $G$. For a theory, once the symmetry group is known, the fields transform under a representation of the group, which is unitary and finite-dimensional. A representation is simply a set of matrices $t^a$ which satisfy the aforementioned commutation relation, which is referred to as a Lie algebra.

Group Action
In general, if a group $G$ acts on a set $X$, there must exist a map, $\phi : G \times X \to X$, such that for all elements of the set, $x\in X$, we have:


*

*$\phi(I,x)=x$ for the identity $I$

*$\phi(g,\phi(h,x))=\phi(gh,x) \,$  for all $g,h \in G$


We denote $\phi$ the group action; essentially it's what we do to the set we act on. For the case of the $U(1)$ group, the structure constants vanish, and when we act on the field, it takes the form,
$$A_{\mu}\to e^{i\alpha} A_{\mu}$$
if we consider global symmetries, then $\alpha \in \mathbb{R}$. If you expand the $\exp(i\alpha)$ as a Taylor series, you see it reproduces the expression in the first equation; unsurprisingly, it's called the exponential map.

Role of handedness
Consider the Dirac fermion; it transforms under a reducible representation; we can decompose it into two irreducible representations which act only one two-component spinors, $u_{\pm}$ Weyl spinors. They transform the same way under rotations, but oppositely under boosts - one is left-handed, the other right-handed.
A consequence of their different handedness is that they lie in different representations;the $u_{+}$ in $(1/2,0)$ whilst the other within $(0,1/2)$. A Dirac spinor lies within  $(1/2,0) \otimes (0,1/2)$, which is treated as a representation of the double cover of the Lorentz group $SL(2,\mathbb{C})$.
