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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Answer To: Problem 1: Let H be a subgroup of group (G, °). a.) Show that NG(H) / CG(H) is isomorphic to a...

David answered on Dec 20 2021
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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 defin
ition 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 either
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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