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Math 55a Midterm – Solutions
Problem 1. Let G be a finite abelian group. Show that the subset of G consisting of those elements
which have odd order is a subgroup.
Let H ⊂ G be the set of elements of odd order. Observe: an element a has odd order if and only
if there exists an odd integer n such that an = e (the order is then some divisor of n, hence odd).
We show the subgroup axioms.
(1) e ∈ H since e has order 1.
(2) let a, b ∈ H, of odd orders n and m. Since G is abelian, (ab)mn = amn bmn = (am )n (bn )m = e.
The order of ab divides mn (which is odd) hence is odd, so ab ∈ H.
(3) let a ∈ H, of odd order m: then (a−1 )m = (am )−1 = e−1 = e, so a−1 has odd order (in fact
equal to m), a−1 ∈ H.
Hence H is a subgroup of G.
Problem 2. Give examples of the following subgroups in the symmetric group S5 , or explain why
none exists.
(a) a cyclic subgroup of order 6;
(b) a non-cyclic subgroup of order 6;
(c) a subgroup of order 7;
(d) a subgroup of order 8.
(a) The permutation (123)(45) has order 6, so the subgroup it generates (which consists of all
(123)k (45)l for k ∈ Z/3, l ∈ Z/2) is a cyclic subgroup of order 6.
(b) The subgroup of permutations such that σ(4) = 4 and σ(5) = 5 is isomorphic to S3 (acting by
permutations on {1, 2, 3} ⊂ {1, 2, 3, 4, 5}).
(c) This is not possible since S5 has order 5! = 120, and 7 does not divide 120.
(d) The dihedral group D4 is a subgroup of order 8 of S4 (by looking at how symmetries of a
square permute the vertices), hence of S5 (setting σ(5) = 5 in all cases). (Equivalently: this is the
subgroup of S5 generated by (1234) and (13)).
Problem 3. Let G be a group (not necessarily abelian), and assume that there exists a short exact
sequence
ϕ
0 −→ Z/2 −→ G −→ Z/k −→ 0,
i.e. there exists a surjective group homomorphism ϕ : G → Z/k with Ker ϕ ' Z/2.
(a) Show that the nontrivial element of Ker ϕ is in the center Z(G), i.e. commutes with every
element of G.
(b) Show that, if g ∈ G is mapped by ϕ to the generator 1 ∈ Z/k, then the order of g in G is either
k or 2k.
(c) Assume k is odd. Show that G is isomorphic to the cyclic group Z/2k. (Hint: use (a) and (b)).
(d) Does the same conclusion (G ' Z/2k) hold if k is even? (give a proof or a counterexample).
(a) Ker ϕ = {e, h} ' Z/2 is a normal subgroup of G, so for each a ∈ G the left and right cosets
a{e, h} = {a, ah} and {e, h}a = {a, ha} must coincide; therefore ah = ha for all a ∈ G.
(b) Let g ∈ G be an element such that ϕ(g) = 1 ∈ Z/k, and let m be the order of g in G, so
g m = e. Then ϕ(g m ) = mϕ(g) = m · 1 = 0, so k divides m. Since |G| = 2k (since |G/ Ker ϕ| = k
and | Ker ϕ| = 2), we conclude that m is either k or 2k.
1
(c) Let g ∈ G be as in part (b), and let g 0 = hg (which is the other element of G mapping to 1).
By part (b), g and g 0 each have order either k or 2k. If g k = e, then (g 0 )k = (hg)k = hk g k = h 6= e,
since k is odd and h commutes with g and has order 2. So g and g 0 cannot both have order k, one
of them has order 2k and this shows that G is cyclic.
(d) The result is false for k even: in fact, G could be either Z/(2k) or Z/2 × Z/k, which is not
cyclic for even k (since ka = 0 for all a ∈ Z/2 × Z/k).
Problem 4. Let φ : V → W be a linear map of vector spaces over a field k. Show that if Ker(φ)
and Im(φ) are both finite-dimensional then V is finite-dimensional.
We need to find a finite set of vectors which span V . Let {u1 , . . . , um } be vectors which span Ker(φ),
and let {w1 . . . , wn } be vectors which span Im(φ). Choose v1 , . . . , vn ∈ V so that wi = φ(vi ). We
claim that {u1 , . . . , um , v1 , . . . , vn } span V .
Indeed, given
v ∈ V ,P
φ(v) ∈ Im(φ) = span(w
P
P 1 , . . . , wn ) so there exist a1 , . . . , an ∈ k such that
φ(v) =
aiP
wi = φ( ai vi ). Thus
v
−
P
P ai vi is in Ker(φ), and can be expressed as a linear
combination
bj uj . Hence v =
ai vi + bj uj ∈ span(u1 , . . . , um , v1 , . . . , vn ).
Problem 5. Let V be a finite-dimensional vector space over the field k, and f, g : V → k two
linear maps. Show that if Ker(f ) ⊂ Ker(g) then there exists λ ∈ k such that g = λ f .
Let dim V = n. If f = 0 then g must be zero as well, and the statement holds; so assume f is
not zero. Then f is surjective, so dim Ker(f ) = n − 1. Let v1 , . . . , vn−1 be a basis for Ker(f ), and
complete it to a basis v1 , . . . , vn for V . Then f (vi ) = g(v
. , n − 1. Setting f (vn ) =
Pni ) = 0 for i = 1, . . P
a 6= 0 and g(vP
ci vi ) = cn f (vn ) = acn
n ) = b, we find that for any vector v =
i=1 ci vi , f (v) = f (
and g(v) = g( ci vi ) = cn g(vn ) = bcn , so g(v) = ab f (v). Hence g = ab f .
(More conceptual solution: Let W = Ker(f ). We have seen that {φ ∈ Hom(V, k) | W ⊂ Ker φ} is
isomorphic to Hom((V /W ), k), which has dimension dim(V /W ) = dim(V ) − dim(W ) = rank f ≤ 1.
So any element of this subspace (in particular, g) is a scalar multiple of f .)
Problem 6. Let V be an n-dimensional vector space over a field k, and f, g : V → V two linear
operators such that f ◦ g − g ◦ f = g.
(a) Show that g maps eigenvectors of f to eigenvectors of f (or to the zero vector).
(b) Show that, if k = C, then g cannot be invertible.
(c) Give an example showing that, if k is a finite field, then g can be invertible.
(a) Let v be an eigenvector of f , f (v) = λv. Then f (g(v)) = g(f (v)) + g(v) = g(λv) + g(v) =
(λ + 1)g(v). So, if g(v) 6= 0 then g(v) is an eigenvector of f , with eigenvalue λ + 1.
(b) If k = C (algebraically closed) then f must have an eigenvector v with some eigenvalue λ. If
g is invartible, then v, g(v), g 2 (v), . . . , g n (v) are all nonzero, and by part (a) they are eigenvectors
of f with eigenvalues λ, λ + 1, λ + 2, . . . , λ + n. However, eigenvectors for distinct eigenvalues must
be linearly independent; this gives a contradiction to the assumption that dim V = n.
(c) Let k = Fp for p prime, and V = k p , with its standard basis which we denote by e0 , . . . , ep−1 .
Take f and g to be the linear operators such that f (ej ) = jej for j = 0, . . . , p − 1, and g(ej ) = ej+1
(where the subscripts are taken mod p, i.e. g(ep−1 ) = e0 ). Then g is clearly invertible, and
f ◦ g(ej ) − g ◦ f (ej ) = f (ej+1 ) − g(jej ) = (j + 1)ej+1 − jej+1 = ej+1 = g(ej ). Eg for p = 3:


 

 

0 0 0
0 0 1
0 0 1
0 0 0
0 0 1
0 1 0 1 0 0 − 1 0 0 0 1 0 = 1 0 0
0 0 2
0 1 0
0 1 0
0 0 2
0 1 0
2
mod 3.