3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $\{(x,y,z) \in \mathbb{C}^3: x + y + z = 0\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.