3.4 Example: D₄ to S₄

Let $H$ be the subgroup of $G=S_4$ isomorphic to $D_4$, obtained by labeling the vertices of a square 1, …, 4 and letting $D_4$ act on them. In other words, $H$ is the image of the injective homomorphism $D_4\to S_4$ sending

$$r\mapsto (1234),s\mapsto (12)(34).$$

Recall that $D_4$ has five irreducible representations with characters as shown:

while recall that $S_4$ has character table

We want to first find out how the conjugacy classes of $S_4$ intersect with $D_4$. The result is as follows, writing $C(g)$ for the conjugacy class of $C(g)$ in $S_4$ or $D_4$.

We use this and the character formula to determine ${\mathrm{Ind}}_H^G{\varphi }_{+}$. The factor $|G|/|H|$ is constant, equal to 3. We obtain:

Decomposing this character, we see that

$${\mathrm{Ind}}_H^G{\varphi }_{+}=ϵ+\psi .$$

But note that we did not need to know the character table of $S_4$ to find the induced character.

We check that this is consistent with Frobenius reciprocity: the restriction of $ϵ$ to $D_4$ is ${\varphi }_{+}$, so

$$⟨{\mathrm{Res}}_H^Gϵ,{\varphi }_{+}⟩=⟨ϵ,{\mathrm{Ind}}_H^G{\varphi }_{+}⟩=1$$

as required. The restriction of $\psi$ to $D_4$ is $\mathrm{𝟙}+{\varphi }_{+}$ so

$$⟨{\mathrm{Res}}_H^G\psi ,{\varphi }_{+}⟩=⟨\psi ,{\mathrm{Ind}}_H^G{\varphi }_{+}⟩=1$$

(we could also check that the restrictions of the other irreducible characters of $S_4$ to $D_4$ do not contain ${\varphi }_{+}$).