Irreducible representation of $S_4$

Question

Could one please point out an irreducible representation of degree 2 of the group $S_4$. Thank you.

I was thinking about mapping $S(4)$ onto $S(3)$ and playing with $S(3)$. Isn't it too complicated? — Jorg, Mar 12 '13 at 21:35
Give it a try. In fact, I don't know, but maybe it works out nicely? — draks ..., Mar 12 '13 at 21:36
Good idea! So can yopu find a 2-dimensional representation of $S_3$? — Derek Holt, Mar 13 '13 at 08:34
Check out this answer. Probably has been done elsewhere on our site as well. — Jyrki Lahtonen, Mar 15 '13 at 04:54

score 3 · Answer 1 · answered Jul 13 '15 at 15:11

Let $\rho^{std}$ be the standard rep. of $S_3$ on $V$ (The dimension of $V$ is equal to $2$). Let $\phi: S_4 \rightarrow S_4/C_2^2$ be a surjective group homomorphism, and $\Omega: S_4/C_2^2 \rightarrow S_3$ a group isomorphism.

Claim: The pullback representation:

\begin{align} p: S_4 &\rightarrow GL(V)\\ g &\mapsto \rho^{std}(\Omega\circ \phi(g)) \end{align}

is an irreducible rep. of $S_4$ on $V$.

Proof: The homomorphic property of $p$ follows directly from the homomorphic property of $\rho^{std}$, $\phi$ and $\Omega$ therefore $p$ is a representation on $V$. Let $W$ be a non trivial invarinat subspace of $V$ under $p$:

$$p(g)w = \rho^{std}(\Omega\circ \phi(g)) w = \rho^{std}(h) w \in W$$

Because $\Omega$ and $\phi$ are surjective this holds for all $h \in S_3$. Thus we get the contradiction that $W$ is a non trivial invarinat subspace of $V$ under $\rho^{std}$.

To now prove the assumptions I made we calculate the Quotient group: $S_4/C_2^2$ and we will see that it is isomorphic to $S_3$. After some calculation we find the Elements of $S_4/C_2^2$:

\begin{align} [C_1] &= \{ (e), (12)(34), (13)(24), (14)(23) \}\\ [C_2] &= \{ (12), (34), (1324), (1423) \}\\ [C_3] &= \{ (13), (24), (1234), (1432) \}\\ [C_4] &= \{ (14), (23), (1243), (1342) \}\\ [C_5] &= \{ (123), (134), (243), (142) \}\\ [C_6] &= \{ (234), (124), (132), (143) \} \end{align}

the natural map $\phi: S_4 \rightarrow S_4/C_2^2$ is obviously surjective and it is a group homomorphsim as the Quotient group of a Conjugacy class is again a group. Now we only have to show that $S_4/C_2^2$ is isomorphic to $S_3$. We define the map

\begin{align} \Omega : S_4/C_2^2 &\rightarrow S_3\\ [C_1] &\mapsto (e)\\ [C_2] &\mapsto (12)\\ [C_3] &\mapsto (13)\\ [C_4] &\mapsto (23)\\ [C_5] &\mapsto (123)\\ [C_6] &\mapsto (132) \end{align}

which is homomorphic (some more calculations) and obviously bijective thus a group isomorphism. This concludes the proof.

Irreducible representation of $S_4$

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