13.6.1 Character Table of
has 3 conjugacy classes: , , . The 3 irreducible representations have dimensions satisfying , so .
| | | |
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| 1 | 3 | 2 |
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| (trivial) | 1 | 1 | 1 |
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| (sign) | 1 | | 1 |
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| (standard) | 2 | 0 | |
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The character of is obtained from the permutation representation: , giving , so . Check: , confirming irreducibility.
13.6.2 Character Table of
has 5 conjugacy classes (partitions of 4): (1), transpositions (6), double transpositions (3), 3-cycles (8), 4-cycles (6). The dimension formula with 5 irreducibles gives .
Three known irreducibles: (trivial, dim 1), (sign, dim 1), (standard, dim 3: permutation rep on minus trivial summand). The tensor product (twist standard by sign) has and is irreducible () and distinct from . This gives 4 of the 5; we need one more of dimension 2 (since , and ).
For the missing (dim 2): use . We have . Compute inner products with known characters to find . Then . Check: .
The complete character table:
Note. The representation has , so double transpositions act trivially. The normal subgroup acts trivially on , so factors through . In fact, is the standard representation of “pulled back” to via the quotient map.
13.6.3 Character Table of
has 4 conjugacy classes: (1), (3), (4), (4). Note the 3-cycles split into two classes (since has no odd permutations to conjugate them). The dimension formula gives .
The three 1-dimensional representations correspond to where , and the 3-dimensional irreducible is the restriction of the standard representation of .
Exercise 13.6.1 (Representation theory exercises).
- Decompose the permutation representation of into irreducibles and verify using the character inner product.
- Let be the standard representation of . Decompose , , and into irreducibles.
- Compute the character table of (the Klein four-group). How many irreducible representations are there, and what are their dimensions?
- Using the character table of , verify that (where are as in the table).
- Prove that for any representation of a finite group , .
Show solution
(1) The permutation representation has character : , , . Compute: , , . So .
(2) : values . Inner products: , , . So .
For : . Values: . So , giving .
For : . Values: . So . (This makes sense: the determinant map of a 2-dimensional rep is the sign character.)
(3) is abelian with 4 elements, so it has 4 conjugacy classes and 4 one-dimensional irreducibles. The dual group is . Writing elements as , the 4 characters are for .
(4) : values . Compute with each irreducible to find multiplicities. This yields (verify the character sum matches).
(5) The averaging operator projects onto . Taking traces: . Since is a projection (), its trace equals its rank: .