If you mean (3_1) is Yang tableaux showing the irreducible representation 3 of the permutation group of S_4, the statement 1. holds but 2. does not hold. The multiplication of two triplets of (3_1) can be decomposed as follows: (3_1)*(3_1)=(3_1)+(2_1_1)+(2_2)+(4) [the last term is the singlet]. If you pick siglet (4) from the right-hand-side and multiplied by a triplet (3_1), you get (3_1) as a result. This is why the statement 2. does not hold. If I have misunderstanding for your statements, this answer is meaningless of course.

Yes, Hiroshi Ozaki is quite right (except perhaps for the spelling of Young tableau). The direct product of a singlet with whatever irrep cannot give something else than that irrep back. In particular, a singlet times a triplet must yield a triplet and nothing else.

A4 group has one triplet and three singlet representations. This follows from the fact that 12 elements of A4 can be divided into 4 classes. Number of conjugacy classes is equal to the number of irreducible representations.  If ni are dimensions of representations then sum over n2i=12.  The sum will go from 1 to 4, because there are four representations. The solution is n1=3, n2=1,n3=1,n4=1 because dimensions are always integers. These representations are denoted by 3, 1, 1',1''. The multiplication rule is

• 3x3=3+3+1+1'+1''
• 1x3=3
• 1'x3=3
• 1''x3=3
• 1'x1''=1        

S4 has five conjugacy classes and 24 elements. Sum over n2i=24 has the solution is 1,1,2,3,3. Here the sum goes from 1 to 5 because there are five representations.

You can see MIT OpenCourseWare course number 18-712

There seems to be some confusion as to whether the question was for S_4 or A_4..

Let us instead of S_4 consider A_4, the 12-element rotational subgroup of the full, 24-element, symmetry group S_4 of a tetrahedron that also contains reflections, and is isomorphic to the perm. group of 4 elements (the corners of the tetrahedron).

Then, of course, Biswajoy is right. However, that does not change the conclusion that a singlet (meaning 1, not 1' or 1'') times anything gives the same anything back.

In particular, 1 (or 1' or 1'') times 3 = 3.

However, here there are two distinct ways to combine 3*3 to give 3, where each can give a singlet when multiplied with another 3.

In the model of Ma and Rajsekharan (http://arxiv.org/abs/hep-ph/0106291), left handed lepton doublets form a flavor triplet of A4 (because there are three generations) whereas, right handed leptons of three generations are assigned to 1,1',1'' representations. There are three SU(2)L doublet Higgs fields transforming as a A4 triplet, whereas there is a extra A4 singlet scalar, which is a SU(2)L doublet. Therefore in total there are four Higgs doublets. With this assignment they could write down necessary mass terms. See Eq. 16 

Hello Ananya Mukherjee. I'm grad I could help. I hope that your work is going well.

Actually I'm ashamed of myself that I typed "Yang tableaux" for the correct spelling of Young tableau. I really appreciate Bo Söderberg for pointing out the misspelling.