Math 502: Abstract Algebra

Homework 3

Lawrence Tyler Rush

<me@tylerlogic.com>

January 5, 2014

http://coursework.tylerlogic.com/courses/upenn/math502/homework03

1

(a) Show that (ℤ∕nℤ)^× is isomorphic to Aut(ℤ∕nℤ).

Any element of Aut(ℤ∕nℤ) will map the identity element to itself because it is a homomorphism. Therefore, since ℤ∕nℤ is cyclic, any automorphism will simply permute the non-identity elements. Hence for any φ ∈ Aut(ℤ∕nℤ), φ(1) = i(1) = i for some i ∈{1,2,…,n}. However, because φ is a homomorphism, this completely dictates the mapping by φ of 2, 3,… n, i.e.

φ(2) = (φ (1))2 = 2i - ( - )3 -- φ(3) = φ (1) = 3i . .. ( - ) -- φ (n-) = φ (1) n = ni

Hence, by defining φ_i to be the automorphism of ℤ∕nℤ that maps 1 to i, we can then see that

Aut(ℤ∕nℤ ) = {φi|gcd(i,n) = 1}

Note that we need the stipulation of i,n being coprime because if j is not coprime to n, then we would be able to find a 0 < k < n such that kj = n, i.e. j wouldn’t generate the group, and therefore φ_j would not be a bijection.

Since (ℤ∕nℤ)^× are the units of ℤ∕nℤ, i.e. the equivalence classes of all the coprime numbers between 0 and n, then the above implies that this is in one-to-one correspondence with Aut(ℤ∕nℤ). The bijection is ϕ : (ℤ∕nℤ)^×→ Aut(ℤ∕nℤ) defined by

- i ↦→ φi

(b) Is (ℤ∕5ℤ)^× a cyclic group?

Yes it is cyclic; it is ⟨2⟩.

-0 2 = 1 mod 25 21 = 2 mod 25 22 = 4 mod 25 -3 2 = 3 mod 25

(c) Extra Credit: Is (ℤ∕25ℤ)^× a cyclic group?

Yes it is cyclic; it is ⟨2⟩, since 5, 10, 15, and 20 are all coprime to 25 (and therefore their equivalence classes are not in (ℤ∕25ℤ)^×

-0 21 = 1 mod 25 2 = 2 mod 25 22 = 4 mod 25 23 = 8 mod 25 -4 25 = 16 mod 25 2 = 7 mod 25 26 = 14 mod 25 27 = 3 mod 25 -8 29 = 6 mod 25 2 = 12 mod 25 210 = 24 mod 25 211 = 23 mod 25 -12 213 = 21 mod 25 2 = 17 mod 25 214 = 9 mod 25 215 = 18 mod 25 -16 217 = 11 mod 25 2 = 22 mod 25 218 = 19 mod 25 219 = 13 mod 25

(d) Extra Credit

2

Let T be the set of diagonal matrices in GL₂(ℝ).

(a) Centralizer of T in GL_n(ℝ)

For n = 2 Because of the following

( ) ( ) ( )-1 ( ) ( ) ( ) a b x a b = ---1-- ax by d - b = --1--- adx - bcy ab(y- x) c d y c d ad- bc cx dy - c a ad- bc cd(x - y) - bcx + ady

we know that for the above to be equal to ( )
x
y , because the determinant ad-bc needs to be non-zero, that a and d need to both be zero and bc = -1, or b and c need to both be zero and ad = 1. Hence,

{ ( ) || } ⋃ { ( ) || } ZGL2(ℝ)(T ) = a 1 ||a ⁄= 0 1 b ||b ⁄= 0 ∕a - ∕b

For n = 3

(b) Normalizer of T in GL_n(ℝ)

For n = 2 For the following to be in T,

( ) ( ) ( ) ( ) ( ) ( ) a b x a b -1 ---1-- ax by d - b --1--- adx - bcy ab(y- x) c d y c d = ad- bc cx dy - c a = ad- bc cd(x - y) - bcx + ady

either a and d need to both be zero, or b and c need to both be zero, but exclusively so because the determinant, ad-bc, needs to be nonzero. Hence we have that

{( a b )|| } NGL2(ℝ)(T ) = b d ||b = c = 0 or a = d = 0

For n = 3

(c) Extra Credit

(d) Extra Credit

3

(a)

(b) Extra Credit

(c) Extra Credit

4 Classify all groups G where Aut(G) = {id_G}

Such a G must be abelian If G were not abelian, then there would be at least one non-identity element, g, outside of the center of G. The map σ_g : G → G defined by σ_g(h) = ghg^-1, i.e. conjugation by g, would then be an automorphism of G since there must exist at least one h ∈ G such that ghg^-1≠h as g ⁄∈ Z(G). Hence σ_g is a non-identity map contained in Aut(G).

5

(a) The trivial subspace and ℝ² are the only subspaces closed under left action by GL₂(ℝ)

It’s clear that the trivial subspace and ℝ² are both closed subspaces under action by GL₂(ℝ). The only other subspaces of ℝ² are 1-dimensional, i.e. lines of ℝ². So simply a rotation, by say π, like

( cos π - sinπ ) sin π cosπ

will take a vector in a 1-dimensional subspace outside of the subspace.

(b)

Functions are stable under action of the identity element. The first property of an action is satisfied by the following

( ) 1 ⋅f(x,y) = f(1x + 0y,0x + 1y) = f(x,y) 1

The “associative” property of an action. Let A,B ∈ GL₂(ℝ) with

( a b) ( m n ) A = c d and B = q r

Because this

( ) A ⋅B ⋅f(x,y) = A⋅ m n ⋅f(x,y) q r ( a b ) = c d ⋅f(mx +qy,nx + ry) = f(m (ax+ cy)+ q(bx+ dy),n(ax+ cy)+ r(bx + dy)) ( ) = f (ma + qb)x+ (mc + qd)y,(na +rb)x+ (nc+ rd)y

is the same as

( ) ( ) (AB )⋅f(x,y) = a b m n ⋅f(x,y) ( c d q r ) am + bq an+ br = cm + dq cn + dr ⋅f(x,y) = f((am + bq)x+ (cm + dq)y,(an + br)x +(cn +dr)y) ( ) = f (am + qb)x+ (mc + qd)y,(na + rb)x +(nc +rd)y

then the “associative” property of an action holds for this action.

(c) Extra Credit

6 Is “is a normal subgroup” a transitive relation?

The “normality” relation is not transitive. In the group D₂₍₈₎ = D₁₆, we have ⟨s,r²⟩⊴ D₁₆ and ⟨s,r⁴⟩⊴⟨s,r²⟩, however, ⟨s,r⁴⟩⁄⊴ D₁₆.

The subgroup ⟨s,r²⟩ is normal in D₁₆ by

k j ℓ2i k j -1 k j ℓ2i- jk k ℓ-j 2i- jk k+ℓ2(i-j) k 2k+ℓ2(j-i) ℓ 2(j-i) 2 (s r)(sr )(s r ) = s r sr r s = s sr r r s = s r s = s r = sr ∈ ⟨s,r ⟩

and ⟨s,r⁴⟩ is normal in ⟨s,r²⟩ by

k 2j ℓ 4i k2j -1 k 2j ℓ4i- 2j k k+ ℓ-2j 4i-2j k k+ℓ 4(i-j)k 2k+ℓ 4(j-i) ℓ 4(j-i) 4 (sr )(s r )(s r ) = s r sr r s = s r r r s = s r s = s r = s r ∈ ⟨s,r ⟩

but the following demonstrates that ⟨s,r⁴⟩⁄⊴ D₁₆

r(sr4)r-1 = rsr3 = sr-1r3 = sr2 ⁄∈ ⟨s,r4⟩