An Introduction to Character Theory J.R. McHugh March 21, 2018 J.R. McHugh An Introduction to Character Theory March 21, 2018 1 / 41 References Abstract Algebra, Third Edition. David S. Dummit and Richard M. Foote. 2004 John Wiley and Sons, Inc. (See chapters 18 and 19.) Character Theory of Finite Groups. I. Martin Isaacs. 1976 Academic Press. A Course in Finite Group Representation Theory. Peter Webb. 2016 Cambridge University Press. (A pre-publication version may be found at Peter Webb's site http://www-users.math.umn.edu/~webb/) J.R. McHugh An Introduction to Character Theory March 21, 2018 2 / 41 Two specific symmetries r and s in D8 are described by the diagrams below: Figure: Motivation Motivation Consider the set D8 whose elements are the symmetries of a square. J.R. McHugh An Introduction to Character Theory March 21, 2018 3 / 41 Motivation Motivation Consider the set D8 whose elements are the symmetries of a square. Two specific symmetries r and s in D8 are described by the diagrams below: J.R. McHugh An Introduction to Character Theory March 21, 2018 3 / 41 Figure: Motivation We define a multiplication in the set D8 by the rule: if x and y are symmetries then xy is the symmetry obtained by first performing y and then performing x. For example, the symmetry rs is described by: J.R. McHugh An Introduction to Character Theory March 21, 2018 4 / 41 It is not difficult to see that every symmetry in D8 is equal to a product of i j the form r s . We say that the group D8 is generated by r and s. The symmetries r and s satisfy r 4 = 1, s2 = 1, and srs = r −1. It turns out that these equalities (called relations) completely determine how symmetries multiply together in D8. We thus obtain a presentation for the group D8: 4 2 −1 D8 = hr; sjr = s = 1; srs = r i: Motivation The multiplication in D8 we have defined is associative and D8 has an identity element 1 and inverses. In other words, the set D8 together with our multiplication is a group. J.R. McHugh An Introduction to Character Theory March 21, 2018 5 / 41 The symmetries r and s satisfy r 4 = 1, s2 = 1, and srs = r −1. It turns out that these equalities (called relations) completely determine how symmetries multiply together in D8. We thus obtain a presentation for the group D8: 4 2 −1 D8 = hr; sjr = s = 1; srs = r i: Motivation The multiplication in D8 we have defined is associative and D8 has an identity element 1 and inverses. In other words, the set D8 together with our multiplication is a group. It is not difficult to see that every symmetry in D8 is equal to a product of i j the form r s . We say that the group D8 is generated by r and s. J.R. McHugh An Introduction to Character Theory March 21, 2018 5 / 41 Motivation The multiplication in D8 we have defined is associative and D8 has an identity element 1 and inverses. In other words, the set D8 together with our multiplication is a group. It is not difficult to see that every symmetry in D8 is equal to a product of i j the form r s . We say that the group D8 is generated by r and s. The symmetries r and s satisfy r 4 = 1, s2 = 1, and srs = r −1. It turns out that these equalities (called relations) completely determine how symmetries multiply together in D8. We thus obtain a presentation for the group D8: 4 2 −1 D8 = hr; sjr = s = 1; srs = r i: J.R. McHugh An Introduction to Character Theory March 21, 2018 5 / 41 Question If we begin with the abstract definition of D8 as a group, how can we recover the square-symmetry interpretation for its elements? Here's one way. Begin by noting that a symmetry of the square permutes its vertices and is uniquely determined by how it permutes the vertices. We then have a one-to-one map X : D8 ! S4: (e.g., X (r) = (1 2 3 4) and X (s) = (2 4).) The map X is known as a permutation representation and is in fact a group homomorphism. Motivation This presentation allows us to define the group \abstractly," i.e., without any reference to the original square and its symmetries. J.R. McHugh An Introduction to Character Theory March 21, 2018 6 / 41 Here's one way. Begin by noting that a symmetry of the square permutes its vertices and is uniquely determined by how it permutes the vertices. We then have a one-to-one map X : D8 ! S4: (e.g., X (r) = (1 2 3 4) and X (s) = (2 4).) The map X is known as a permutation representation and is in fact a group homomorphism. Motivation This presentation allows us to define the group \abstractly," i.e., without any reference to the original square and its symmetries. Question If we begin with the abstract definition of D8 as a group, how can we recover the square-symmetry interpretation for its elements? J.R. McHugh An Introduction to Character Theory March 21, 2018 6 / 41 Motivation This presentation allows us to define the group \abstractly," i.e., without any reference to the original square and its symmetries. Question If we begin with the abstract definition of D8 as a group, how can we recover the square-symmetry interpretation for its elements? Here's one way. Begin by noting that a symmetry of the square permutes its vertices and is uniquely determined by how it permutes the vertices. We then have a one-to-one map X : D8 ! S4: (e.g., X (r) = (1 2 3 4) and X (s) = (2 4).) The map X is known as a permutation representation and is in fact a group homomorphism. J.R. McHugh An Introduction to Character Theory March 21, 2018 6 / 41 When a group action of G on Ω exists, we say G acts on Ω. For example, if V is a vector space over the field F then the multiplicative group F × = F n f0g acts on V since c · (d · v) = cd · v and 1 · v = v for all c; d 2 F and v 2 V (these are axioms built into the definition of a vector space). Another example: any group G acts on itself by left multiplication. Motivation Another way is to use a group action. Definition Let G be a group and Ω a set. A group action of G on Ω is a map from G × Ω to Ω written g · ! for g 2 G, ! 2 Ω that satisfies both g · (h · !) = gh · ! and 1 · ! = ! for all g; h 2 G and ! 2 Ω. J.R. McHugh An Introduction to Character Theory March 21, 2018 7 / 41 Motivation Another way is to use a group action. Definition Let G be a group and Ω a set. A group action of G on Ω is a map from G × Ω to Ω written g · ! for g 2 G, ! 2 Ω that satisfies both g · (h · !) = gh · ! and 1 · ! = ! for all g; h 2 G and ! 2 Ω. When a group action of G on Ω exists, we say G acts on Ω. For example, if V is a vector space over the field F then the multiplicative group F × = F n f0g acts on V since c · (d · v) = cd · v and 1 · v = v for all c; d 2 F and v 2 V (these are axioms built into the definition of a vector space). Another example: any group G acts on itself by left multiplication. J.R. McHugh An Introduction to Character Theory March 21, 2018 7 / 41 The fact that this defines an actual group action hinges on the fact that X is a homomorphism: g · (h · !) = X (g)(X (h)(!)) = (X (g) ◦ X (h))(!) = X (gh)(!) = gh · ! and 1 · ! = X (1)(!) = !: We can interpret each element of D8 as a symmetry of the square via this group action. Motivation Let X : D8 ! S4 be the permutation representation of D8 from before. We can define a group action of D8 on Ω = f1; 2; 3; 4g by g · ! = X (g)(!) for g 2 D8 and ! 2 Ω. For example, r · 1 = 2 and s · 2 = 4. J.R. McHugh An Introduction to Character Theory March 21, 2018 8 / 41 We can interpret each element of D8 as a symmetry of the square via this group action. Motivation Let X : D8 ! S4 be the permutation representation of D8 from before. We can define a group action of D8 on Ω = f1; 2; 3; 4g by g · ! = X (g)(!) for g 2 D8 and ! 2 Ω. For example, r · 1 = 2 and s · 2 = 4. The fact that this defines an actual group action hinges on the fact that X is a homomorphism: g · (h · !) = X (g)(X (h)(!)) = (X (g) ◦ X (h))(!) = X (gh)(!) = gh · ! and 1 · ! = X (1)(!) = !: J.R. McHugh An Introduction to Character Theory March 21, 2018 8 / 41 Motivation Let X : D8 ! S4 be the permutation representation of D8 from before. We can define a group action of D8 on Ω = f1; 2; 3; 4g by g · ! = X (g)(!) for g 2 D8 and ! 2 Ω. For example, r · 1 = 2 and s · 2 = 4. The fact that this defines an actual group action hinges on the fact that X is a homomorphism: g · (h · !) = X (g)(X (h)(!)) = (X (g) ◦ X (h))(!) = X (gh)(!) = gh · ! and 1 · ! = X (1)(!) = !: We can interpret each element of D8 as a symmetry of the square via this group action.