Let $\chi$ be 2-dimensional complex character of the group $ S_3 $. Prove that $\chi$ is irreducible character iff $\chi((123))=-1$.

There is hint in my book: " Use the Maschke's theorem and the properties of the commutator subgroup"

I try. Assume that $\chi$ is reducible character of the group $ S_3. $ Using the Maschke's theorem we have $\chi=\chi_1+\chi_2,$ as $\chi_1, \chi_2$ are $1$-dimensional complex character of the group $ S_3 $. But $$\chi_1(123)= \chi_2(123)=-1$$

I can not prove " if $\chi$ is irreducible character then $\chi((123))=-1$"

I know that $S_3 '=A_3.$ But I can not use it.

I am sorry for my English. Thanks a lot in advance for any help!

  • $\begingroup$ Do you mean $1$ dimensional when you say $-$ dimensional complex characters of group $S_3$? $\endgroup$ – user87543 Jan 27 '14 at 10:20
  • 2
    $\begingroup$ Why can you not use $S_3'=A_3$? Any $1$-dimensional character $\chi_1$ has $S_3'=A_3$ in its kernel, so $\chi_1((1,2,3))=1$, $\chi((1,2,3))=2$. $\endgroup$ – Derek Holt Jan 27 '14 at 10:52
  • $\begingroup$ @Praphulla Koushik. Oops I fixed it. Thank you! $\endgroup$ – nadia-liza Jan 27 '14 at 13:33
  • $\begingroup$ @ Derek Holt Thank you! I understood. I cannot prove follow statements " if χ is irreducible character then χ((123))=−1" I edited. $\endgroup$ – nadia-liza Jan 27 '14 at 13:51

To prove that any $2$-dimensional irreducible complex character $\chi$ has $\chi((123)) = -1$, there are many approaches.

  • You could use the orthogonality relations: $\sum_{\psi \in \operatorname{Irr}(G)} |\psi(g)|^2 = |C_G(g)|$. Hence $\chi((123))^2 = 1$, and since $\chi$ has trivial kernel it follows that $\chi((123)) = -1$.

  • You could note that if $\rho: S_3 \rightarrow \operatorname{GL}_2(\mathbb{C})$ affords $\chi$, then $\rho((123))$ is similar to a diagonal matrix $\pmatrix{\alpha & 0 \\ 0 & \beta}$, where $\alpha^3 = \beta^3 = 1$. Then $\chi((123)) = \alpha + \beta$ and $\chi((123)) = \chi((132)) = \alpha^2 + \beta^2$, you can see by looking at the possible values of $\alpha$ and $\beta$ (3rd roots of unity) that we must have $\chi((123)) = -1$.

  • You could note that there is a unique $2$-dimensional irreducible character, hence it suffices to find an irreducible $2$-dimensional representation and calculate $\chi((123))$ directly. Since $S_3$ is isomorphic to the symmetry group of an equilateral triangle, a natural linear action on the triangle on a plane should work. It is a faithful representation of dimension $2$, hence irreducible since $S_3$ is nonabelian.

  • An irreducible representation can also be found from the permutation action of $S_3$ on $\mathbb{C}^3$ (permutation of coordinates), then the vector $(1,1,1)$ is invariant and has a $2$-dimensional irreducible complement $\{(x,y,z) \in \mathbb{C}^3: x + y + z = 0\}$.


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