If one tries to verify the construction of the standard model, one has to find a Lagrangian that is invariant under $U(1)\times SU(2) \times SU(3)$. While it seems kind of logic that the Lagrangian should be invariant under rotation, it seems a little bit arbitrary that it should be invariant under $SU(2)$ and $SU(3)$ transformations. I understand that the $SU(2)$ group is constructed through the Pauli matrices, but still, this alone seems like a poor argument.
Gauge Invariance – Importance of SU(2) and SU(3) Gauge Invariance in the Standard Model
gauge-theorygroup-theorylagrangian-formalismquantum-field-theorystandard-model
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Well, if something is invariant under the action of some group it means that every element leaves it invariant. In particular, every element that belongs to some subgroup. Also, you can get an obvious representation of the subgroup by restriction. If $\rho: G \to {\rm Aut}(V)$ is a representation then so is $\rho_{H}: H \to {\rm Aut}(V)$ for $H < G$. Naturally, this need not be irreducible anymore (obvious example being representation of a trivial subgroup which is reducible to $N$ components).
Well, they surely do because one can take $T\oplus \dots \oplus T$ where $T$ is trivial one dimensional representation. Let's pretend that you meant irreducible representations so that this question is actually non-trivial. Groups of $A_l$ type (with $l = N-1$) are of rank $l$. This means that finite-dimensional representations (as is the case for the irreducible representations of compact $\mathbf {SU}$ forms of these groups) are indexed by $l$ non-negative integers $\{\lambda_i\}_{i=1}^l$. E.g. for $l=1$ (${\mathbf {SU}(2)}$) this integer is connected to spin by $\lambda_1 = 2s$ (and dimension of these representations is $\lambda_1 +1 = 2s+1$). So we are interested in determining dimensions of the irreducible representations indexed by $\lambda$. This is not so straight-forward; in principle one can determine these dimensions by looking at the weight space of the given representation and counting the number of ways one can arrive at any given weight starting from highest weight and applying annihilation operators.
Instead, one can exploit the fact that $\mathbf {SL}$ is basically the same thing as $ \mathbf {GL}$ and use Schur-Weyl duality between $\mathbf {GL}(N,\mathbb C)$ and $\mathfrak{S}_k$ (i.e. the symmetry group of $k$-element set) which states that the representation of the above groups on $(\mathbf C^n)^{\otimes k}$ (with obvious actions) decomposes as sum over partitions $\lambda \in Par(k,N)$ (e.g. $(3,2,1) \in Par(6,3)$) of products of irreducible representations $F^{\lambda} \otimes G^{\lambda}$ (with $F$ resp. $G$ being a representation of $\mathbf {GL}(N,\mathbb C)$, resp. $\mathfrak{S}_k$). The dimensions of these representations can then be computed by fancy combinatorial formulas derived from Young tableaux (which enter due to partitions, of course). Basically one writes down some numbers related to $n$ and number of rows under and columns to the right into each box of the tableaux and multiplies all those numbers together. As for whether every integer greater or equal to $N$ can be obtained in this way, I don't know. I suppose by shear amount of these tableaux this should be true but I'll come back to check it later.
We do not start from the assumption that the Lagrangian "should" be invariant under gauge transformations. This assumption is often made because global symmetries are seen as more natural than local symmetries and so writers try to motivate gauge theory by "making the global symmetry local", but this is actually nonsense. Why would we want a local symmetry just because there's a global one? Do we have some fetish for symmetries so that we want to make the most symmetrical theory possible? One can derive gauge theory this way but as a physical motivation, this is a red herring.
The actual point is not that we "want" gauge symmetry, but that it is forced upon us when we want to describe massless vector bosons in quantum field theory. As I also allude to in this answer of mine, every massless vector boson is necessarily described by a gauge field. A Lagrangian gauge theory is equivalently a Hamiltonian constrained theory - either way, the number of independent degrees of freedom that are physically meaningful is less that the naive count, since we identify physical states related by gauge transformations.
The true physical motivation for gauge theories is not "we want local symmetries because symmetries are neat". It's "we want to describe a world with photons in it and that can only covariantly be done with a gauge theory".
A non-quantum motivation of gauge theory can also be given: If you write down the Lagrangian of free electromagnetism, motivated because its equations of motion are the Maxwell equations, not because we like gauge symmetry, then you find it comes naturally with a $\mathrm{U}(1)$ gauge symmetry, corresponding to the well-known fact that adding a gradient to the vector 4-potential is physically irrelevant. Now, if you want to couple other fields to this free electromagnetism, you need to make the additional terms gauge invariant, too, else the theory is no longer "electromagnetism coupled to something else" in any meaningful sense since suddenly adding gradients can change the physics. Once again, gauge symmetry is something one discovers after physically motivating the Lagrangian from something else, not some sort of a priori assumption we put in.
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As far as we know, within the standard framework of quantum field theory it is arbitrary that the Standard Model has $\mathrm{U}(1)\times \mathrm{SU}(2)\times\mathrm{SU}(3)$ as its gauge group. We choose this group because it predicts the correct particle content and interactions that we observe e.g. in colliders, not because of some compelling theoretical reason to choose this group over any other.
Quantum field theories with many other gauge groups are consistent and perfectly valid theories - they just don't seem to describe our universe (though not all choices see consistent, see "anomalies", "Landau poles", etc.).