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I have an exercise to show that $\mathbb{Q}S_3 \cong \mathbb{Q} \oplus\ \mathbb{Q}\ \oplus\ M_2(\mathbb{Q}) $ , where $M_2(\mathbb{Q})$ is ring of $2$ by $2$ rational matrices and $\mathbb{Q}S_3$ is group ring.

How can I establish ring isomorphism between them. After several attempts, I cannot find a suitable isomorphism from $\mathbb{Q}S_3 \to \mathbb{Q} \oplus\ \mathbb{Q}\ \oplus\ M_2(\mathbb{Q})$.

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  • $\begingroup$ Do you know a little group representation theory ? $\endgroup$ – Captain Lama Mar 28 '16 at 20:54
  • $\begingroup$ @CaptainLama I know a little, but I was looking for explicit isomorphism and how can we go about finding one in such cases. I do not mind if it involves some representation theory. $\endgroup$ – Bhaskar Vashishth Mar 28 '16 at 20:56
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    $\begingroup$ You can look up Maschke's theorem and Artin–Wedderburn theorem. $\endgroup$ – Ben Mar 28 '16 at 21:00
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First let's treat the case of $\mathbb{C}[G]$. for any finite group $G$, $\mathbb{C}[G]\simeq \prod_i Hom_\mathbb{C}(V_i)$ where the $(V_i,\rho_i)$ are the non-isomorphic irreducible representations of $G$, and the isomorphism is given by $g\mapsto (\rho_i(g))_i$.

Now if $G=S_3$, this gives $\mathbb{C}[S_3] \simeq M_2(\mathbb{C})\times \mathbb{C}\times \mathbb{C}$ since $S_3$ has $3$ irreducible representations : the standard one, the trivial one, and the signature. But you may check that they all are defined over $\mathbb{Q}$, meaning that $\rho(g)\in M_r(\mathbb{Q})$ for all these representations.

So this gives you by restriction $\mathbb{Q}[S_3] \simeq M_2(\mathbb{Q})\times \mathbb{Q}\times \mathbb{Q}$, and the isomorphism is $g\mapsto (\rho(g),1,\varepsilon(g))$ where $\rho:S_3\to M_2(\mathbb{Q})$ is the standard representation and $\varepsilon(g)$ is the signature.

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  • $\begingroup$ I'm not sure how you get the $\rho$. The standard representation of $S_3$ in $M_2(\mathbb R)$ - using that $S_3$ is a dihedral group - is not in $M_2(\mathbb Q)$ because the coefficients are in $\mathbb Q[\sqrt{3}]$. $\endgroup$ – Thomas Andrews Mar 28 '16 at 21:31
  • $\begingroup$ Well, you can certainly choose a basis such that the coefficients are not in $\mathbb{Q}$, but for the standard choice, the coefficients are in $\mathbb{Z}$, like every representation of any $S_n$. The standard representation is the quotient of the canonical permutation representation (defined over $\mathbb{Q}$ like any permutation representation) by the trivial subrepresentation (also defined over $\mathbb{Q}$). $\endgroup$ – Captain Lama Mar 28 '16 at 21:38
  • $\begingroup$ @CaptainLama As $\mathbb{C}[S_3] \simeq M_2(\mathbb{C})\times \mathbb{C}\times \mathbb{C}$, so in this case also the the same isomorphism works? $\endgroup$ – Bhaskar Vashishth Apr 18 '16 at 21:15

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