Finding the inverse matrix responsible for base change in the Z3 Group

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SUMMARY

The discussion focuses on finding the inverse matrix responsible for a base change between the subgroup of S3 and the group Z3. The permutations involved are σ_{1}=(1)(2)(3), σ_{5}=(1,2,3), and σ_{6}=(1,3,2), which form a subgroup isomorphic to Z3 represented as {1,ω,ω^{2}}. The matrix S is derived from the linear combination of basis vectors, and the determinant is calculated as 3ω(ω-1). The participants discuss the method for finding the inverse matrix S^{-1}, including the use of the transpose of the matrix of cofactors and the determinant.

PREREQUISITES
  • Understanding of group theory, specifically the properties of permutations and isomorphic groups.
  • Familiarity with linear algebra concepts, including matrix operations and determinants.
  • Knowledge of complex numbers, particularly the properties of ω where ω^3=1.
  • Experience with calculating matrix inverses using the cofactor method.
NEXT STEPS
  • Study the properties of the symmetric group S3 and its subgroups.
  • Learn about the process of finding the inverse of a matrix using the cofactor method.
  • Explore the implications of complex numbers in group theory, particularly roots of unity.
  • Investigate linear transformations and their representations in matrix form.
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Mathematicians, students studying abstract algebra, and anyone interested in group theory and linear algebra applications.

Dixanadu
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Homework Statement


Hey guys,
So I have the following permutations, which are a subgroup of S3:
σ_{1}=(1)(2)(3), σ_{5}=(1,2,3), σ_{6}=(1,3,2)
This is isomorphic to Z3, which can be written as {1,ω,ω^{2}}

Next, we have the basis for the subgroup of S3:
e_{i}=e_{1},e_{2},e_{3}

And we also have the basis for the group Z3, which is a linear combination of the basis vectors of S3:
E_{k}=(e_{1}+e_{2}+e_{3}), (e_{1}+ωe_{2}+ω^{2}e_{3}), (e_{1}<br /> +ω^{2}e_{2}+ωe_{3})

I have to find the matrix (and its inverse) which is responsible for the following base change:
E_{k}=S_{jk}e_{j}

Homework Equations


Dont think there are any.

The Attempt at a Solution


So I think I've found the matrix S, the problem is that how do I find S^{-1}. I get the following for S, even though I'm not sure its right:
http://imageshack.com/a/img401/9545/h213.jpg

I tried using the regular matrix rules for finding the inverse (transpose of the matrix of cofactors divided by the determinant) but it doesn't seem to work.

Can you guys help me out?

thanks in advance.
 
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You realize that, by definition of \omega, \omega^3=1, right? The determinant of the matrix is 3\omega(\omega- 1) and the "transpose of the matrix of cofactors" is
\begin{pmatrix}\omega(\omega- 1) &amp; \omega(\omega- 1) &amp; \omega(\omega- 1) \\ \omega(\omega- 1) &amp; \omega- 1 &amp; (\omega- 1)(\omega+ 1) \\ \omega(\omega- 1) &amp; (\omega- 1)(\omega+ 1) &amp; \omega- 1\end{pmatrix}

For example, calculating the "1,1" cofactor, we compute
\left|\begin{array}{cc}\omega &amp; \omega^2 \\ \omega^2 &amp; \omega \end{array}\right|= \omega^2- \omega^4= \omega^2- (\omega^3)\omega= \omega^2- \omega= \omega(\omega- 1)
 
Hi HallsofIvy,
I get all elements of the matrix the same as you, apart from the ones [2,3] and [3,2] - [rows,columns]. You have ω^{2}-1 from which you take the difference of two squares. However, isn't the place value of that element in the matrix negative? which means that you should have -(ω^{2}-1)=(1+ω)(1-ω)? if that's correct then I'm stuck again because it doesn't factor into the determinant anymore!
 

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