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Determination of character table of alternating group:A5

1,682 bytes added, 00:12, 15 March 2017
Putting in the remaining representations
| second irreducible three-dimensional representation || 3 || ? || ? || ? || ?
|}
 
=== Filling in the column of order two elements ===
Let's fill in the second column (corresponding to the conjugacy class <math>(1,2)(3,4)</math>). The element has order 2, so its image under any representation has eigenvalues <math>\pm 1</math>. Therefore, the character value is a sum of values that are <math>\pm 1</math>, with the number of summands being the degree. Therefore:
| second irreducible three-dimensional representation || 3 || -1 || ? || ? || ?
|}
 
=== Filling in the five-dimensional representation ===
Next, let us try to fill in the irreducible five-dimensional representation. We know that it is the only five-dimensional representation, so it equals its complex conjugate. Therefore, it has a real-valued character. We also know that the two conjugacy classes <math>(1,2,3,4,5)</math> and <math>(1,2,3,5,4)</math> are in the same automorphism class. Since the representation is the only one of its degree, it is automorphism-invariant, hence the character must be equal on these two classes. We therefore get:
|-
| second irreducible three-dimensional representation || 3 || -1 || ? || ? || ?
|}
 
=== Filling in the column of order three ===
 
Examine the missing entries <math>a</math> and <math>b</math>:
 
{| class="sortable" border="1"
! Representation/conjugacy class representative and size !! <math>()</math> (size 1) !! <math>(1,2)(3,4)</math> (size 15) !! <math>(1,2,3)</math> (size 20) !! <math>(1,2,3,4,5)</math> (size 12) !! <math>(1,2,3,5,4)</math> (size 12)
|-
| trivial || 1 || 1 || 1 || 1 || 1
|-
| restriction of standard || 4 || 0 || 1 || -1 || -1
|-
| irreducible five-dimensional representation || 5 || 1 || -1 || 0 || 0
|-
| first irreducible three-dimensional representation || 3 || -1 || <math>a</math> || ? || ?
|-
| second irreducible three-dimensional representation || 3 || -1 || <math>b</math> || ? || ?
|}
 
By the [[column orthogonality theorem]] on the column containing these entries, we have:
 
<math>1^2 + 1^2 + (-1)^2 + a^ 2 + b^2 = 60/20 = 3</math>
 
Thus:
 
<math>a^2 + b^2 = 0</math>
 
This forces <math>a = b = 0</math>>
 
We thus get:
 
 
{| class="sortable" border="1"
! Representation/conjugacy class representative and size !! <math>()</math> (size 1) !! <math>(1,2)(3,4)</math> (size 15) !! <math>(1,2,3)</math> (size 20) !! <math>(1,2,3,4,5)</math> (size 12) !! <math>(1,2,3,5,4)</math> (size 12)
|-
| trivial || 1 || 1 || 1 || 1 || 1
|-
| restriction of standard || 4 || 0 || 1 || -1 || -1
|-
| irreducible five-dimensional representation || 5 || 1 || -1 || 0 || 0
|-
| first irreducible three-dimensional representation || 3 || -1 || 0 || ? || ?
|-
| second irreducible three-dimensional representation || 3 || -1 || 0 || ? || ?
|}
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