# All different 2d representations of S3

I'm trying to find all non-isomorphic 2 dimensional representations of the symmetric group, $$S_3$$. This is all my own work, so I shan't be surprised if it's horrendously wrong.

I understand that two representations cannot be isomorphic if they have different kernels and images, so I'm thinking that one "good" way to find a set of potential representations would be to look at representations of various "faithfulnesses".

Since $$\ker(\rho) = \{1\} \iff \rho \text{ is injective}$$, I'll look at $$\rho: S_3 \rightarrow \mathbb{R^2}$$ with varying kernels.

1. Firstly, a 2D representation, $$\rho_1$$ is simply $$S_3$$ acting as $$D_3$$ (the dihedral group with 6 elements) on $$\mathbb{R^2}$$, this has $$\ker(\rho_1) = \{1\}$$.
2. Look at $$\rho_2$$, which is just a representation with a non trivial kernel. E.g. $$\rho_2((12)) = Id$$, this would mean $$\rho_2((123)) = \rho_2((13)) \rho_2((12)) = \rho_2((13))$$, but these are cycles of differing order, so the only way to preserve the group structure is if $$\rho_2((123)) = \rho_2((13)) = Id$$. Of course this then means that $$\rho_2$$ is the trivial representation
3. If $$\rho_3$$ has a non-trivial kernel but, say, $$(123) \in \ker(\rho_3)$$. This then implies $$(132) \in \ker(\rho_3)$$, we're left to define $$\rho_3$$ on the two cycles s.t. they still have order 2 and multiply to give the identity. I think this means they all must be equal and one of three matrices:$$\begin{pmatrix}-1&0\\0 & 1\end{pmatrix}$$ Or, $$\begin{pmatrix}1&0\\0 & -1\end{pmatrix}$$ Or, $$\begin{pmatrix}-1&0\\0 & -1\end{pmatrix}$$
4. Should the above be true, then it is clear that there is no longer enough room for a still "smaller" representation, therefore there are 4 different representations total.

## Questions

• Is the above working correct?

• I'm unsure as to whether I can claim that: Any two faithful representations of the same finite group, $$G$$, on the same vector space, $$V$$, are $$G$$-isomorphic?

• Can I generate even more representations by considering a different 2 dimensional vector space, which is not $$\mathbb{R^2}$$? If not, how can this be proven?

## Thoughts

1. My thinking is that "Any two faithful representations of the same finite group, $$G$$, on the same vector space, $$V$$, are $$G$$-isomorphic" is true:

We'd just need that $$T: (\rho_a, V) \rightarrow (\rho_b, V)$$ has $$T(\rho_a(g)v) = \rho_b(g)T(v)$$ But there is a bijection between the set of $$\{\rho_a(g) : g \in G \}$$ and $$\{\rho_b(g) : g \in G \}$$ and $$G$$ being finite means that after relabelling they should be exactly the same representation, with $$T$$, a trivial intertwiner.

2. I'm also not sure if there is anything similar if the representations aren't faithful , I have tacitly assumed a similar result for 2. and 3. to guarantee that these are all the representations

• The notation $\rho: S_3\to\Bbb R^2$ is wrong. The representation is a homomorphism $\rho:S_3\to GL_2(\Bbb R)$.
– anon
Oct 31, 2019 at 21:06

(1) If two representations of a group on a vector space are isomorphic, then their images are conjugate subgroups of the general linear group, but can be distinct subgroups.

(2) Kernels are normal subgroups. So if a group homomorphism $$\rho: S_3\to\mathrm{GL}_2(\mathbb{R})$$ has a kernel containing an element like $$(12)$$, then it contains all of its conjugates, which are also transpositions, and transpositions generate all of $$S_3$$ so the kernel is all of $$S_3$$ and the representation is trivial.

(3) The only (proper nontrivial) normal subgroup of $$S_3$$ is the alternating subgroup $$A_3$$, which happens to be the cyclic group $$C_3$$ generated by the $$3$$-cycle $$(123)$$. Then $$A=\rho((12))$$ must have order $$2$$, as you note. However, there are infinitely many matrices of order $$2$$. For instance, reflections across a line are order $$2$$, and there are infinitely many lines. In fact, every such $$A$$ acts like a reflection across a line in some coordinates.
Factor $$A^2v=v$$ as $$(A+I)(A-I)v=0$$ or $$(A-I)(A+I)v=0$$ for all $$v$$, hence there exist nonzero eigenvectors $$x$$ and $$y$$ (in the range of $$A+I$$ and $$A-I$$) such that $$Ax=x$$ and $$Ay=-y$$. Thus, through a change of coordinates, $$A$$ is conjugate to the diagonal matrix $$\mathrm{diag}(1,-1)$$. So all the matrices $$A$$ of order $$2$$ are conjugate, and the corresponding representations are all equivalent.
(4) A more advanced fact in representation theory states the sum of the (complex) irreps' squared dimensions is the size of the group squared. Here, we have two $$1$$D reps (trivial and sign) and the $$2$$D rep (dihedral), and since $$1^2+1^2+2^2=3!^2$$ (where $$|S_3|=3!$$), there can't be any more reps. (We're talking about real irreps, but in this case the sum of the squared dimensions is $$\le |G|^2$$, so the logic still works.)
(5) Two faithful representations of a finite group on a vector space are not necessarily isomorphic. Even if their images are the same subgroup (not just conjugate), they aren't necessarily isomorphic. A way to produce counterexamples is using outer automorphisms. If $$G$$ is a finite group, $$\rho:G\to GL(V)$$ a representation, and there is an outer automorphism $$\alpha:G\to G$$, then $$\rho\circ\alpha$$ is inequivalent to $$\rho$$, but has the same image.