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The 3-dimensional representation of the group S3 can be constructed by introducing a vector $(a,b,c)$ and permute its component by matrix multiplication.

For example, the representation for the operation $(23):(a,b,c)\rightarrow(a,c,b)$ is

$ D(23)=\left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}\right) $

and so forth.

The exercise is to prove this representation is reducible. The hint tells me to find a common eigenvector for all 6 matrices which is just $(1,1,1)$. How do I proceed from here? Any help is appreciated.

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Here's another way to prove it's reducible, although it may depend on stuff you haven't learned yet. The order of the group is the sum of the squares of the degrees of the irreducible representations. So a group of order 6 can't have an irreducible representation of degree 3; $3^2\gt6$.

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You asked: The exercise is to prove this representation is reducible. The hint tells me to find a common eigenvector for all 6 matrices which is just $(1,1,1)$. How do I proceed from here? Any help is appreciated.

Representation is irreducible if and only if the corresponding $FG$-module has no non-trivial $FG$-submodule. If you have managed to find a common eigenvector $v$, then you have $vg=\lambda_g v$ for each $g\in G$; which implies that $\operatorname{span}(v)$ is an $FG$-submodule.

In short: Looking for 1-dimensional submodules is the same thing as looking for common eigenvectors. If your the whole $FG$-module has dimension 3, it suffices to find out whether it has a 1-dimensional submodule, in order to decide whether it is irreducible or not.

It is perhaps worth mentioning that the same approach would work for any $S_n$ and that this representation is called permutation representation of $S_n$. Another interesting fact is that the permutation representation can be decomposed into this trivial representation and an irreducible representation of degree $n-1$. We have a question about this on this site; link to this MO thread is given there in comments.

Note: My answer is more-or-less the same as Benjalim's answer (which is deleted at the moment, so it is visible only for 10k+ users), with the exception that my answer uses modules and his answer avoids modules and uses only representations. (Both approaches, $FG$-modules and representations, are equivalent in the sense that we can get module from a representation and vice-versa. Hence we can describe properties of representation using the properties of the corresponding $FG$-module.)

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