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Problem 1: Let H be a subgroup of group (G, °). a.) Show that NG(H) / CG(H) is isomorphic to a subgroup of Aut(H). b.) Show that CG(H) = NG(H) if |H| = 2. Problem 2: Let G be a group of order |G| =...

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Problem 1: Let H be a subgroup of group (G, °).
a.) Show that NG(H) / CG(H) is isomorphic to a subgroup of Aut(H).
b.) Show that CG(H) = NG(H) if |H| = 2.
Problem 2: Let G be a group of order |G| = 992 = 25 · 31.
a.) Show that at least one of its Sylow p-subgroups is normal.
b.) If G is Abelian, find all prime-power decompositions and invariant decompositions.
Problem 3: Prove that if G / Z(G) is cyclic, then G is Abelian.
Problem 4: Let G be a group of order 99.
a.) Prove that there exists a subgroup of H of order 3. (Should be a one-sentence proof.)
b.) Prove there is a unique subgroup K of G such that K / H ˜ Z3 and G / K ˜ Z11.
Problem 5: Characterize those integers n such that the only Abelian groups of order n are cyclic.
Problem 6: Express U(165) as an internal direct product of proper subgroups in 4 different ways.
Problem 7: Suppose (R, +, °) is a ring in which a ° a = 0 implies a = 0. Show that R has no nonzero nilpotent elements.
Problem 8: Let A and B be ideals of a ring (R, +, °). If A n B = {0}, show that a ° b = 0 whenever a ? A and b ? B.
Problem 9: Show that x4 + 1 is irreducible over Q but reducible over R. What is the smallest extension field containing Q for which x4+ 1 is reducible?
Notation:
NG(H) denotes the normalizer of H in G.
CG(H) denotes the centralizer of H in G.
Aut(H) denotes the class of all automorphisms of H.
Z(G) denotes the center of G.
U(165) denotes the group of units for 165.
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Solve each problem: Problem 1: Let H be a subgroup of group (G, °). a.) Show that NG(H) / CG(H) is isomorphic to a subgroup of Aut(H). b.) Show that CG(H) = NG(H) if |H| = 2. Problem 2: Let G be a group of order |G| = 992 = 25 · 31. a.) Show that at least one of its Sylow p-subgroups is normal. b.) If G is Abelian, find all prime-power decompositions and invariant decompositions. Problem 3: Prove that if G / Z(G) is cyclic, then G is Abelian. Problem 4: Let G be a group of order 99. a.) Prove that there exists a subgroup of H of order 3. (Should be a one-sentence proof.) b.) Prove there is a unique subgroup K of G such that K / H ˜ Z3 and G / K ˜ Z11. Problem 5: Characterize those integers n such that the only Abelian groups of order n are cyclic. Problem 6: Express U(165) as an internal direct product of proper subgroups in 4 different ways. Problem 7: Suppose (R, +, °) is a ring in which a ° a = 0 implies a = 0. Show that R has no nonzero nilpotent elements. Problem 8: Let A and B be ideals of a ring (R, +, °). If A n B = {0}, show that a ° b = 0 whenever a ? A and b ? B. Problem 9: Show that x4 + 1 is irreducible over Q but reducible over R. What is the smallest extension field containing Q for which x4+ 1 is reducible? Notation: NG(H) denotes the normalizer of H in G. CG(H) denotes the centralizer of H in G. Aut(H) denotes the class of all automorphisms of H. Z(G) denotes the center of G. U(165) denotes the group of units for 165.

Answered Same Day Dec 20, 2021

Solution

David answered on Dec 20 2021
126 Votes
1. (a) We can define a map χ : NG(H)→ Aut(H) by sending g to the automorphism σg : H → H obtained
from conjugation by g. By definition of normalizer we have σg(h) = ghg
−1 ∈ H whenever h ∈ H and
g ∈ NG(H), so σg is well defined. Then g ∈ Ker(χ) if and only if σg(h) = h for all h ∈ H, if and only
if gh = hg for all h ∈ H, i.e, if and only if g ∈ CG(H). By the First Isomorphism Theorem we have
NG(H)/CG(H) ∼= Im(χ), and note that Im(χ) is a subgroup of Aut(H).
1. (b) Since |H| = 2, therefore H is abelian and hence all elements commute with each other and so CG(H) =
NG(H).
2. (a) The number of 31-Sylow subgroups must divide 31 × 32 and be congruent to 1 modulo 31; so eithe
there is a single 31-Sylow subgroup (in which case the group is not simple), or there are thirty two
31-Sylow subgroups.
If there are thirty two 31-Sylow subgroups, then since any two distinct ones must intersect trivially (the
groups are cyclic of prime order, so the only proper subgroup is trivial), they account for 32(31−1)+1
elements of G.
That means that there are 32 × 31 − 32 × 30 = 32 elements whose order is not 31. Since a 2-Sylow
subgroup must contain 32 elements, there are only enough elements left over for a *single* 2-Sylow
subgroup, which must therefore be normal.
So G will have either a...
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