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→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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