Download Math 6310 Homework Solutions by Cesar Lozano: Group Theory and more Assignments Algebra in PDF only on Docsity!
Math 6310
Homework 4^1
C´esar Lozano
Problem 1 Let G be a group which contains a subgroup of index 11 and an element of order
- Show that G is not simple.
Solution. To see this we need to take into account two cases
|G| = ∞, & |G| 6 = ∞.
Case: |G| = ∞. This case was solved within the assignment number two:
Let G be an infinite group and H a subgroup of finite index. Show that G has a normal subgroup K of finite index with K < H.
Let K < G be the subgroup of index 11. Such an assumption implies K 6 = e. Therefore, remembering assignment 2, G is not simple. Case: |G| 6 = ∞. Here we are going to argue by contradiction. Since [G : H] = 11, let us consider the set of cosets G/H = {a 1 H,... a 11 H},
and an action of the group G on this set
G × G/H −→ G/H (g, aiH) 7 −→ (gaiH). Since gaiH = aj H for some j, this action is giving us a permutation σg(i) = j, which only dependents on the element g ∈ G. Indeed, we have gotten a map
f : G, −→ S 11 g 7 −→ σg.
Now suppose G is a simple group, therefore, kerf = e and hence G ∼= f (G) < S 11. From now on, for simplicity in the notation let’s write f (G) = G < S 11. Then there is an element σ ∈ S 11 of order 30. Furthermore, G < A 11 if G is not contained in A 11 , which means by the second theorem of isomorphism G must have a subgroup, namely G ∩ A 11 , which has index
- Then G < A 11.
(^1) October 28 ,2008.
Now the problem follow from the observation that if σ ∈ A 11 has order 30, then there must be an element σ′^ ∈ A 11 such that |σ′| = 15^2. However, considering σ′^ as a product of transpositions
σ′^15 = σ^151 · · · σ^15 r = σ 1 · · · σr, we realize σ′^ cannot have order 15.
Problem 2 Show that the automorphism group of Zn is isomorphic to Z× n. Show explicitly how each element of Z× n acts on Zn.
Solution. Let x be a generator of the cyclic group Zn. If f ∈ Aut(Zn), then f (x) = xa^ for some a ∈ Z. Observe that the integer a determines the automorphism f , so we can denote it as fa. Particularly fa is a homomorphism, hence x and xa^ must have the same order. Indeed (n, a)=1 and for every a relatively prime to n, the map x 7 → xa^ is an automorphism of Zn. Therefore, we have a surjective map
F : Aut(Zn) −→ Z× n fa 7 −→ a(mod n) This map F is indeed a group homomorphism because
fa ◦ fb = fa(xb) = xab^ = fab,
this for all fa, fb ∈ Aut(Zn). Moreover, F (fa ◦ fb) = F (fab) = ab(mod n) = F (fa)F (fb). Finally, F is clearly injective, hence it is an isomorphism.
Problem 3 Let G ⊂ H abelian groups. Show that the following condition are equivalents:
- There exist a homomorphism f from H to G such that the restriction of f to G is the identity
- There exists s subgroup K of H such that H = G ⊕ K
Solution. Suppose the first statement, there is a map f : H → G. Observe this mao is surjective. Then consider the short exact sequence
0 //kerf //H (^) f //G //
i } } 0
where the map i is the inclusion one. Note this sequence splits since i ◦ f = id. Therefore, if exists a subgroup K such that H = G ⊕ K.
(^2) Cauchy’s Theorem and the fact that |ab| = |a||b| (a,b) for^ a, b^ ∈^ G
Problem 6 Show that Q/Z is isomorphic to ⊕ p
Z(p∞), where the direct sum is over all prime
numbers p.
Solution. Observe that Q/Z = Q ∩ [0, 1], since Z[1/p] = {ab ∈ Q|b = pm}, and hence
Z(p∞) = Z[1/p]/Z,
Let us define a map between these two spaces as follows
Q/Z −→ ⊕pZ(p∞) x = (^) pβ 1 a 1 ···p βnn 7 −→^ ap
−β 2 2 · · ·^ p −βn n ⊕ · · · ⊕^ ap
−β 1 1 · · ·^ p
−β(n−1) (n−1).
where pβ 11 · · · pβ nn is the prime decomposition of the number b ∈ Z and x = ab ∈ Q. This definition makes this map an isomorphism between
Q/Z ∼= ⊕ p
Z(p∞).
as desired.
Problem 7 Let C be a category. let C × C be the category whose objects are ordered pairs (A, B) of objects of C and whose morphisms are ordered pairs of morphism of C with com- ponent wise composition. Let D be the functor form C to C × C which sends an object A to (A, A) and the morphism α to (α, α). Show that a coproduct in C defines a left adjoint of D.
Solution. Let suppose there exists the coproduct in the category C. Now Let us define a functor (^) ∐ : C × C −→ C,
which in objects Ob(C) has the following definition:
C D^ //C × C
∐ / / C
A D^ //(A, A)
(A, B)
∐ / / A ∐^ B
We are going to use doted arrow in order to distinguish the action of the functors. The functor F in arrows is defined as follows:
C D^ //C × C
∐ / / C
(P : A → B) ^ D^ ___//(P, P )
(P :(A,B)→(C,D))
∐ _ _ _ / /(A ∐^ B→C ∐^ D)
Thus, we want to show that
HomC (A
B, C) ∼= HomC×C ((A, B), D(C)).
Let us see why this is true.
HomC (A
B, C)
Φ −→ HomC×C ((A, B), D(C))
(g :A ∐^ B→C) (PA ◦ g, Pb ◦ g) (g :A ∐^ B→C) (PA, PB )
where the maps PA, PB and h as specified in the following diagram
A
PA (^) //
PA◦g
? ?? ??
??? ?? ?? ?? ?? ?? ??
A
B
∃!^ g
^
(^)
^ B
oo^ PB
PB ◦g
C
Since the map g above is unique, the map Ψ is a bijection.
The University of Utah
Department of Mathematics lozano@math.utah.edu
