# Changes

## Determination of character table of alternating group:A5

, 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 $(1,2)(3,4)$). The element has order 2, so its image under any representation has eigenvalues $\pm 1$. Therefore, the character value is a sum of values that are $\pm 1$, 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 $(1,2,3,4,5)$ and $(1,2,3,5,4)$ 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 $a$ and $b$:

{| 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 || $a$ || ? || ?
|-
| second irreducible three-dimensional representation || 3 || -1 || $b$ || ? || ?
|}

By the [[column orthogonality theorem]] on the column containing these entries, we have:

$1^2 + 1^2 + (-1)^2 + a^ 2 + b^2 = 60/20 = 3$

Thus:

$a^2 + b^2 = 0$

This forces $a = b = 0$>

We thus get:

{| 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 || ? || ?
|-
| second irreducible three-dimensional representation || 3 || -1 || 0 || ? || ?
|}