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Determination of character table of alternating group:A5

2,175 bytes added, 00:20, 15 March 2017
Filling in the column of order three
<math>a^2 + b^2 = 0</math>
This forces <math>a = b = 0</math>>.
We thus get:
| second irreducible three-dimensional representation || 3 || -1 || 0 || ? || ?
|}
 
=== Filling in the four remaining entries (for the three-dimensional representations) ===
 
Let's label them:
 
{| 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 || <math>j</math> || <math>k</math>
|-
| second irreducible three-dimensional representation || 3 || -1 || 0 || <math>l</math> || <math>m</math>
|}
 
By the [[character orthogonality theorem]], we get:
 
<math>(1)(1)(3) + (15)(1)(-1) + (20)(1)(0) + (12)(1)(j) + (12)(1)(k) = 0</math>
 
and:
 
<math>(1)(3)^2 + (15)(1)^2 + (20)(0)^2 + (12)(j)^2 + (12)(k)^2 = 60</math>
 
This simplifies to:
 
<math>j + k = -1</math>
 
and:
 
<math>j^2 + k^2 = 3</math>
 
Plugging into a single equation, we get:
 
<math>j^2 + (-1 - j)^2 = 3</math>
 
This becomes:
 
<math>2j^2 + 2j - 2 = 0</math>
 
We solve to get <math>j = (1 \pm \sqrt{5})/2</math>. Note that since the same constraints apply to the other character as well, the two solutions represent the two possible characters. Plugging in, we get the full character table:
 
{| 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 || 5 || 1 || -1 || 0 || 0
|-
| one irreducible constituent of restriction of exterior square of standard || 3 || -1 || 0 || <math>(\sqrt{5} +1)/2</math> || <math>(-\sqrt{5} + 1)/2</math>
|-
| other irreducible constituent of restriction of exterior square of standard || 3 || -1 || 0 || <math>(-\sqrt{5} + 1)/2</math> || <math>(\sqrt{5} + 1)/2</math>
|}
 
This simplif
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