[image]

with strict inclusions, such that each H_i is a maximal strict normal subgroup of H_i+1. Equivalently, a composition series is a subnormal series such that each factor group H_i+1 / H_i is simple. The factor groups are called composition factors. Source

Notes

Prove...

IF (G ·) is finite with Ø ≠ S ⊂ G, S closed under ·

THEN (S ·) is a group.

By two-step subgroup test we only need show S closed under inversion:

For b ∈ S , m = |b| we have (b^m-1)b = b(b^m-1) b^m = e. So b^m-1 is the inverse of b and it's in S since S is closed under ·.

Define...

Two-step subgroup test

A nonempty subset of a group is itself a group if the subset is closed under the operation as well as under the taking of inverses.

Wiki

Prove...

A nonempty subset of a group is itself a group if the subset is closed under the operation as well as under the taking of inverses. (Two-step subgroup test)

Let S be such that

Ø ≠ S ⊂ G,

S closed under · ,

x ∈ S → x^-1 ∈ S .

Associativity is inherited from G since S ⊂ G. We only need show that e ∈ S:

Ø ≠ S → ∃ b∈ S → b^-1 ∈ S → e = bb^-1 ∈ S.

Example...

Give an example to show (aH)(bH) = (ab)H doesn't necessarily hold if H is not a normal subgroup.

Take H = {(1,2), ()} in S₃.

a = (1,3), b = (2,3), ab = (1,3,2)

aH = {(1,2,3) (1,3)}

bH = {(1,3,2) (2,3)}

(ab)H = {(1,3) (1,3,2)}

But

(aH)(bH)

= {(1,2,3) (1,3)} {(1,3,2) (2,3)}

= {() (1,2) (2,3) (1,3,2)}

Prove...

∀ a,b ∈ G [ (ab)² = a²b² ] → G is abelian

ab = eabe = (a^-1a)ab(bb^-1)

= a^-1(aabb)b^-1 = a^-1(abab)b^-1

= (a^-1a)ba(bb^-1) = ebae = ba

Prove...

In a finite group G, ∃N | ∀a∈G, a^N = e .

Each element has finite order so we only need take N to be the LCM of the orders.

To show each element element has finite order consider

e = a⁰, a¹, a², a³, ..., a^|G|.

This sequence has |G| + 1 terms but only |G| possible values. So by pigeonhole principle: a^j = a^k , for some j < k. Thus a^k-j = e.

Prove...

If a^-1 = a, ∀a ∈ G then G is Abelian.

ab = a^-1b^-1 = (ba)^-1 = ba

Prove ...

A group of order ≤ 4 is Abelian.

If the group is non-Abelian then there are distinct elements say x,y such that xy ≠ yx.

That means x ≠ e ≠ y, which in turn means xy ≠ x ≠ yx and xy ≠ y ≠ yx. (If xy = x then we would have y = e by cancellation.)

Also, xy ≠ e ≠ yx. (If xy = e then x and y would be inverses and commute.)

Therefore, xy and yx must be distinct 4th and 5th elements.

(In fact, any non-Abelian group must ≥ 6 elements.)

Give an Example...

Infinite Field of Finite Characteristic > 0

One very important example of an infinite field of characteristic p is Fp(T)={f/g | f,g in Fp[T], g!= 0}, the rational functions in the indeterminate T with coefficients in Fp (the symbol Fp is just a synonym for Z/pZ). In other words, these are ratios of polynomials in Fp[T]; this is the same construction as the one we use to make Q from Z. The field Fp(T) is infinite because, for example, it contains 1, T, T2, …, and it is of characteristic p because it contains Fp (alternatively, because the kernel of the unique ring homomorphism Z→Fp(T) is pZ.)

source

Prove...

In an ID: prime → irreducible.

a = bc, both non units

→ a | bc

→ a | b or a | c, e.g. WLOG a | b

→ b = ak

→ a = akc

→ 1 = kc

→ c is a unit ><

Prove...

Group homomorphism maps identity to identity

φ(a)φ(e)= φ(ae) = φ(a) = φ(a)e'

-->  φ(e) = e' by cancelation in G'.

Prove...

For any subset H of G and any g ∈ G,

 g ∈ H  iff gH ⊆ H

g ∈ H → gH ⊆ H since H is closed.

gH ⊆ H → g ∈ H since ge ∈ gH (as e ∈ H).

Let S be the set of left cosets of a subgroup H in G.

Show that S partitions G.

Lemma: aH = bH iff a∈bH.

Proof: Suppose aH=bH.

Then a = ae ∈ aH =bH.

Conversely, if a∈bH then a=bh (hence aH ⊆ bH) and b=ah^-1 (hence bH ⊆ aH).

Now suppose aH and bH overlap via c. 

Then c = ah₁ = bh₂. for some h₁, h₂ ∈ H.

Thus a = bh₂h₁^-1 and a∈bH.  

By the Lemma above, we conclude aH = bH.

Thus distinct left cosets are nonoverlapping.

Every x ∈ G belongs to a left coset of H (namely xH) and every left coset of H is a subset of G. 

Thus the union of all left cosets of H is just G .

So the set of left cosets form a partition of G.

Prove

If H and K, normal subgroups of G whose only common element is the identity then the elements of H and K commute. 

Prove...

If C = Z(G) then G/C is cyclic iff G is abelan. 

Assume that G is abelian.

Then C=G so G/C is trivial and therefore cyclic.

Assume G/C is cyclic, i.e. G/C has generator tC.

Then each coset is (tC)ⁱ = tⁱC for some i.

Let x, y be elements of G.

Then x ∈ t^mC and y ∈ tⁿC for some m,n.

i.e. x=t^ma and y=tⁿb for some a,b ∈ C.

Now...

xy

= t^ma tⁿb

= t^mtⁿab

=tⁿt^mba

= tⁿb t^ma

= yx.

Notice we used the fact that a,b ∈ C to freely swap their positions with the other factors.

The Sylow subgroups are of orders 5 and 3² = 9. 

There can be only one Sylow subgroup of order 5 since the only divisor of 45 congruent to 1 (mod 5) is 1.

Similarly, there can be only one Sylow subgroup of order 9 since the only divisor of 45 congruent to 1 (mod 9) is 1.

Call these H₅ and H₉.

Since H₅ and H₉ are unique subgroups of their orders, they must be normal.

H₅ is of prime order so it must be isomorphic to Z5 .

H₉ is isomorphic to either Z₃ x Z₃ or Z₉.

The intersection H3 and H9 is trivial since only 1 divides both 5 and 9.

(H₃)(H₉) is a group since H₃ (alternatively, H₉) is normal.

Moreover, (H₃)(H₉) must equal G since its order is a multiple of both 5 and 9.

Therefore G is the internal direct product of H₅ and H₉.

Since there were two possible structures for H9 we have two possible structures for G:

Z₅ x Z₉ = Z₄₅  or  Z₅ x Z₅ X Z₃ = Z₃ x Z₁₅.

So G is either cyclic of order 45 or the product of two cyclic groups of orders 3 and 15.

Prove

If H U K is a subgroup then H sub K or K sub H

hk must be in H U K so in H or in K. 

If hk in H then hk = h' --> k = h^-1h' So k in H.

Thus K is contained in H.

If hk in K then hk = k' --> h = k^-1k' So h in K.

Thus H is contained in K.

If x is nilpotent, then 1 − x is a unit.

If x is nilpotent, then 1 − x is a unit, because xⁿ = 0 entails

(1 - x)(1 + x + x² + ... + x^n-1) = 1 - xⁿ = 1

More generally,

the sum of a unit element and a nilpotent element

is a unit when they commute.

Let R be a ring. An element x of R is called nilpotent if there exists an integer m ≥ 0 such that x m = 0.

(a) Show that every nilpotent element of R is either zero or a zero-divisor.

(b) Suppose that R is commutative and let x, y ∈ R be nilpotent and r ∈ R arbitrary. Prove that x + y and rx are nilpotent.

(c) Now suppose that R is commutative with an identity and that x ∈ R is nilpotent. Show that 1 + x is a unit and deduce that the sum of a unit and a nilpotent element is a unit. 

(a) Let x ∈ R be nilpotent and let m ≥ 0 be the minimal integer such that x^m = 0. If m = 0 then either x = 0 or 1 = 0 and R is the zero ring (hence x = 0). If m = 1 then x = 0. If m > 1 and x ≠ 0 then 0 = x m = x · x m−1 with m − 1 > 0 and x^m−1 ≠ 0 by minimality of m. Thus x is a zero-divisor.

(b) Suppose that x^m = 0 and yⁿ = 0 and let N be any integer greater than m + n. By commutativity of R, we have the binomial theorem (same proof as usual) so in the expansion for (x + y) N every monomial has the form a_ijxⁱy^j with i + j = N so either i > m or j > n or both. Thus, every monomial term is zero and hence x+y is nilpotent. Again by commutativity, (rx)^m = r^mx^m = 0 so rx is nilpotent.

(c) Suppose x^m = 0.

Then (1+x)(1−x+x² − · · ·+ (−1)^m−1x^m−1 ) = 1, so 1 +x is a unit.

If u is any unit and x is nilpotent, u + x = u(1 + u⁻¹x) is the product of two units (using that u⁻¹x is nilpotent by the above) and hence a unit.

Find conjugacy classes in S₃

S₃ = { () (12)  (13)  (23)  (123)  (132) }

a()a^-1 = () for any a in S₃ so {()} is a class.

(12)(123)(12) = (132)

(13)(123)(13) = (132)

(23)(123)(23) = (132)

So {(123), (132)} is a class

(132)(12)(123) = (13)

(132)(13)(123) = (23)

(132)(23)(123) = (12)

So {(12), (13), (23)} is a class

In general the congugacy classes will correspond to partitions of S_n.  Here we have

3   (rotations)

2 + 1   (pairs)

1 + 1 + 1  (singletons i.e. just identity perm)

To determine the size of a conjugacy class in S_n ,

choose a permutation from the class

represented by its cyclic decompositon.

For each 

 j = 1 to n, let a_j be the number of cycles of length j 

then f_j  is j^aj(a_j)!

Now divide n! by the product of the f_j.

In the case of S₃ and the conjugacy class

{(12), (13), (23)} 

we can choose (12) = (12)(3) as our representative.

f₁ is  1¹(1)! =1

f₂ is  2¹(1)! = 4

f₃ is  3⁰(0)!  = 1

So the size of our conjugacy class is 3!/2 = 3.

Find conjugacy classes in S₄

S₄ = {

()

(12)   (13)   (23)   (14)   (34)  (24)

(12)(34)   (13)(24)    (14)(23)   

(123)   (234)   (243)  (132)  (134)   (143)    (124)   (142) 

(1234)     (1423)   (1243)  (1432)  (1324)    (1342) 

}

The conjugacy classes are the permutations with same structure. 

There are 5 classes since P(4) = 5.

The size of each class is given by n! / (f₁f₂...f_n)

 f_j  is j^aj(a_j)!   with a_j = how many cycles of size j.

For the (ab)(cd) class 

f₁ is  1⁰(0)! = 1 

f₂ is  2²(2)! = 4 

f₃ is  3⁰(0)! = 1 

f₄ is  4⁰(0)! = 1 

So the size of our conjugacy class is 4!/8 = 3.

For the (ab) class 

f₁ is  1²(2)! = 2 

f₂ is  2¹(1)! = 2 

f₃ is  3⁰(0)! = 1 

f₄ is  4⁰(0)! = 1 

So the size of our conjugacy class is 4!/4 = 6.

For the (abc)(d) class 

f₁ is  1¹(1)! = 1 

f₂ is  2⁰(0)! = 1 

f₃ is  3¹(1)! = 3 

f₄ is  4⁰(0)! = 1 

So the size of our conjugacy class is 4!/3 = 8.

For the (abcd) class 

f₁ is  1⁰(0)! = 1 

f₂ is  2⁰(0)! = 1 

f₃ is  3⁰(0)! = 1 

f₄ is  4¹(1)! = 4 

So the size of our conjugacy class is 4!/4 = 6. 

Prove Wilson's Theorem

(p-1)! == -1 (mod p)

Also if (n-1)!== -1 (mod n) then n is prime

Since Z_p is a field,

every nonzero element has a multiplicative inverse.

That means the product will be congruent to 1 (mod p).

Also, because Z_p is a field, x² - 1 has at most two roots. 

That means only 1 an -1 are their own inverses.

All the other elements can be paired up with their inverses.

Taking the product of the pairs gives 1.

Now we can multiply by 1 and p-1 to get -1 (mod p).

So (p-1)! ≡ -1 (mod p).

For the converse?

State and prove Fermat's 2 square theorem.

Fermat's [image] theorem, sometimes called Fermat's two-square theorem or simply "Fermat's theorem," states that a prime number [image] can be represented in an essentially unique manner (up to the order of addends) in the form [image] for integer [image] and [image] iff[image] or [image] (which is a degenerate case with [image]). The theorem was stated by Fermat, but the first published proof was by Euler.

Dedekind's two proofs using Gaussian integers[edit]

Richard Dedekind gave at least two proofs of Fermat's theorem on sums of two squares, both using the arithmetical properties of the Gaussian integers, which are numbers of the form a + bi, where a and b are integers, and i is the square root of −1. One appears in section 27 of his exposition of ideals published in 1877; the second appeared in Supplement XI to Peter Gustav Lejeune Dirichlet's Vorlesungen über Zahlentheorie, and was published in 1894.

1. First proof. If [image] is an odd prime number, then we have [image] in the Gaussian integers. Consequently, writing a Gaussian integer ω = x + iy with x,y ∈ Z and applying the Frobenius automorphism in Z[i]/(p), one finds

[image]

since the automorphism fixes the elements of Z/(p). In the current case, [image] for some integer n, and so in the above expression for ω^p, the exponent (p-1)/2 of -1 is even. Hence the right hand side equals ω, so in this case the Frobenius endomorphism ofZ[i]/(p) is the identity. Kummer had already established that if f ∈ {1,2} is the order of the Frobenius automorphism of Z[i]/(p), then the ideal [image] in Z[i] would be a product of 2/f distinct prime ideals. (In fact, Kummer had established a much more general result for any extension of Z obtained by adjoining a primitive m-th root of unity, where m was any positive integer; this is the case m = 4 of that result.) Therefore the ideal (p) is the product of two different prime ideals in Z[i]. Since the Gaussian integers are a Euclidean domainfor the norm function [image], every ideal is principal and generated by a nonzero element of the ideal of minimal norm. Since the norm is multiplicative, the norm of a generator [image] of one of the ideal factors of (p) must be a strict divisor of [image], so that we must have [image], which gives Fermat's theorem.

Prove

If p = 4k+1

then p-1 is a square (quadratic residue) (mod p).

Take (1)(4k)(2)(4k-1)...(2k)(2k+1) = (p-1)! = -1

(by Wilson's Theorem).

But

(1)(4k)(2)(4k-1)...(2k)(2k+1) 

= 1(-1)2(-2)...(2k)(-2k) = ((2k)!)²

(Note an even number of factors.)

So -1 = c²

Where c = ((p-1)/2)!.

Note Z[sqrt(-5)] is not a principal ideal domain. So the factors may not be principle ideals.

In fact 2 is irreducible but not prime since 2 | (1+sqrt(-5))(1-sqrt(-5))  but 2 doesn't divide either factor.

Suppose (2) = I². 

Try I = (2, 1 + sqrt(-5))