# Changes

## Determination of character table of alternating group:A5

, 00:20, 15 March 2017
Filling in the column of order three
$a^2 + b^2 = 0$
This forces $a = b = 0$>.
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 !! $()$ (size 1) !! $(1,2)(3,4)$ (size 15) !! $(1,2,3)$ (size 20) !! $(1,2,3,4,5)$ (size 12) !! $(1,2,3,5,4)$ (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 || $j$ || $k$
|-
| second irreducible three-dimensional representation || 3 || -1 || 0 || $l$ || $m$
|}

By the [[character orthogonality theorem]], we get:

$(1)(1)(3) + (15)(1)(-1) + (20)(1)(0) + (12)(1)(j) + (12)(1)(k) = 0$

and:

$(1)(3)^2 + (15)(1)^2 + (20)(0)^2 + (12)(j)^2 + (12)(k)^2 = 60$

This simplifies to:

$j + k = -1$

and:

$j^2 + k^2 = 3$

Plugging into a single equation, we get:

$j^2 + (-1 - j)^2 = 3$

This becomes:

$2j^2 + 2j - 2 = 0$

We solve to get $j = (1 \pm \sqrt{5})/2$. 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 !! $()$ (size 1) !! $(1,2)(3,4)$ (size 15) !! $(1,2,3)$ (size 20) !! $(1,2,3,4,5)$ (size 12) !! $(1,2,3,5,4)$ (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 || $(\sqrt{5} +1)/2$ || $(-\sqrt{5} + 1)/2$
|-
| other irreducible constituent of restriction of exterior square of standard || 3 || -1 || 0 || $(-\sqrt{5} + 1)/2$ || $(\sqrt{5} + 1)/2$
|}

This simplif