Rep Theory Flashcards
State and prove Maschke’s Theorem.
Let G a finite group, V a finite dim rep of G over a field k where char(k)=0 or char(k) does not divide |G|. Then if U a subrep of V, there exists another subrep W such that U+W=V.
Proof:
Find the character table of S_4.
Give bounds on the norm (inner product with itself) of a restriction or induction of an irrep. Give a bound on the occurrences of an irreducible character in another based on dimensions.
Both are bounded by the index of the group. ????
- Show that a simple non-abelian simple group has no non-trivial linear reps.
- Show that an order 2 element has integer character congruent to the dimension of the rep mod 4.
- Given the fact that the dimension of a rep divides the order of G, show a simple group has no irred dim 2 reps.
- Show that the number of irred linear reps divides the order of G.
Proof 1: The linear reps are exactly those lifted from G/G’, where G’ is the derived subgroup. Since G’ is normal in G, either G’ is trivial (i.e., G is abelian) or G’=G. Hence G’=G. Hence there is a unique linear rep.
Proof 2: Given a non-trivial linear rep, the kernel is trivial so we identify G as a subgroup of the multiplicative group of a field, hence G is cyclic.
Describe how to find the normal subgroups from the character table. Show that the lifts of Irr(G/N) are exactly the irred reps with kernel N.
Show that an odd group has a single real irreducible character.
Lemma 1: If X irred, 1/|G| sum X(g^2) is 0 if X is not real, +/- 1 if X is real.
Let X be real. Say g^2=h^2. Then g^{2m}=h^{2m}=1, m odd. Then g^m=h^m=1. Therefore g=g^{m+1}=h^{m+1}=h. So the sum over g^2 is the same as the sum of g. Hence =1. Done!
- Show that groups of order p^2 are abelian
- Show that in a non-abelian p-group, the number of linear reps n_1 is divisible by p^2.
- Show that in a p-group the degree d of a rep satisfies d^2 divides [G:Z(G)].
- The dims of irred reps are 1, p or p^2. Since the sum of the squares of the dims is p^2, all reps are 1.
- Note that d divides |G|, hence is a power of p. So is [G:Z(G)]. So it suffices to show d^2 <= [G:Z(G)]. Use |G| = sum |X(g)|^2 >= sum_{g in Z(G)} |X(g)|^2 >= |Z(G)| d^2.
Show that given an irred rep of G, and a subgroup H of G, there is an irred rep of H which, when induced to G, has G occuring as a component once.
- Show that if A <= G is abelian, then any irred rep of G has dimension at most [G:A].
- Show that if an irred rep has dimension [G:A], then
- If X is an irred rep of G, then there is an irred rep of A which when induced to G has X as a component. Note that irred reps of A are linear reps, so when induced to G they have dimension [G:A]. Hence dim X <= [G:A].
- If X has dimension [G:A], then X must equal this induced rep.
Define a topological group and a compact group, and a representation for a topological group.
V must be finite dimensonal. Then we give GL(V) the natural topology inherited from C or R ^{n^2} for n=dimensional V.
What are the 1-d irred characters of S^1? Prove this assuming that all cts group homos R–>S^1 take a certain form.
We assume cts group homos R–>S^1 are of the form x |–> e^(icx).
Define SU(2). What is the maximal torus T in SU(2). Define V_n and explain how it is a rep of SU(2). What is the character of an element of T?
Prove that V_n is irred.
- Note that =C_m <= SU(2) for every m. Note that sigma x^jy^{n-j}=omega^{2j-n} x^j y^{n-j}. Pick m large enough so that these are all distinct eigenvalues. Note that if W <= V_n, then W is a -rep, hence decomposes as a sum of 1-dimensional irred components. Well, we’ve found all of them, so W = Direct sum a_i , a_i =0 or 1.
- But then, note that action by 1/sqrt[2}(1,-1,1,1) on x^j y^{n-j} gives nonzero coeff on x^n. Thus a_n = 1.
- Then, use that matrix again on x^n, to get every a_i=1 for all i. So, done.
Show that the character table is square assuming row orthogonality.
It suffices to show that the only class function orthogonal to all irred reps is 0.
Consider an irred rep V. Consider such a class function f. We then define a G-linear map V–>V via the dot product of the rep with f. By Schur’s plus trace, one sees this map is 0.
Now consider the regular rep of G. The above function is zero on each irred subspace of this rep, hence is 0. Since the above function is zero, it maps e_1 to 0. But then f must be everywhere 0.
Show that if G a group having a ccl of non trivial prime power, then G is not non-abelian simple. Deduce Burnside’s theorem.
Deduce that if A <= G is abelian of prime power index, then G is not non-abelian simple.
- If G non-abelian simple, chi irrep of dim d>1, and O a ccl of size coprime to d, then chi=0 on O.
Proof: aChi(1)+b|O|=1, and Chi(g)/Chi(1)=aChi(g)+b Chi(g)|O|/Chi(1) is an alg integer. Thus Chi(g)/Chi(1)=0, or it has absolute value 1. If it has absolute value 1, then g must act as a scalar, hence the action is not faithful (as G has trivial centre), hence the rep must be the trivial rep. But d>1. So done.
- If |O| has prime power, then G not non-abelian simple.
Proof: Suppose G were. Then, consider the set of non-trivial irreps of G. So they all have dimension at least 2. Either they have dimension div by p, or else chi(g)=0. So, consider column orthogonality. Then 0=1 + sum Chi_i(1) Chi_i(g), so p divides 1. - If G has size p^aq^b, consider Q a sylow-q subgroup. If g != 1 in Z(Q), then C(g) contains Q, thus the ccl of g has size a power of p, and at least p since not everything centralizes g. So done.
- The same proof shows that if A <= G abelian of prime power index, then G not non-abelian simple.
Show that if |G|=2m, m odd, then G has an index two normal subgroup.
Consider the regular rep of G. This injects G into S_{2m}, which then maps to {+/-1} via the sign homomorphism. Pick g an order 2 element. Then note that g has no fixed points, hence is m 2-cycles, hence is odd.
