Term

Definition
[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 


Term
Prove...
IF (G ·) is finite with Ø ≠ S ⊂ G, S closed under ·
THEN (S ·) is a group. 

Definition
By twostep subgroup test we only need show S closed under inversion:
For b ∈ S , m = b we have (b^{m1})b = b(b^{m1}) b^{m} = e. So b^{m1} is the inverse of b and it's in S since S is closed under ·.



Term
Define...
Twostep subgroup test 

Definition
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 


Term
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. (Twostep subgroup test) 

Definition
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.



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

Definition
Take H = {(1,2), ()} in S_{3}.
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)}



Term
Prove...
∀ a,b ∈ G [ (ab)^{2} = a^{2}b^{2} ] → G is abelian 

Definition
ab =^{} eabe = (a^{1}a)ab(bb^{1})
= a^{1}(aabb)b^{1} = a^{1}(abab)b^{1}
= (a^{1}a)ba(bb^{1}) = ebae = ba



Term
Prove...
In a finite group G, ∃N  ∀a∈G, a^{N} = e . 

Definition
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 = a^{0}, a^{1}, a^{2}, a^{3}, ..., 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^{kj } = e. 


Term
Prove...
If a^{1} = a, ∀a ∈ G then G is Abelian. 

Definition
ab = a^{1}b^{1} = (ba)^{1} = ba 


Term
Prove ...
A group of order ≤ 4 is Abelian. 

Definition
If the group is nonAbelian 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 nonAbelian group must ≥ 6 elements.) 


Term
Give an Example...
Infinite Field of Finite Characteristic > 0 

Definition
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 


Term

Definition


Term
Prove...
If ker(φ) = {0} then φ is an injection. 

Definition
φ(a) = φ(b)
→ φ(a)  φ(b) = 0
→ φ(a  b) = 0
→ a  b = 0
→ a = b. 


Term
Prove...
In an ID: prime → irreducible. 

Definition
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 ><



Term
Prove...
Group homomorphism maps identity to identity 

Definition
φ(a)φ(e)= φ(ae) = φ(a) = φ(a)e'
> φ(e) = e' by cancelation in G'. 


Term
Prove...
For any subset H of G and any g ∈ G,
g ∈ H iff gH ⊆ H 

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



Term
Let S be the set of left cosets of a subgroup H in G.
Show that S partitions G. 

Definition
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 = ah_{1} = bh_{2}. for some h_{1}, h_{2} ∈ H.
Thus a = bh_{2}h_{1}^{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.



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

Definition


Term
^{} Prove...
If C = Z(G) then G/C is cyclic iff G is abelan. 

Definition
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} = t^{i}C for some i.
Let x, y be elements of G.
Then x ∈ t^{m}C and y ∈ t^{n}C for some m,n.
i.e. x=t^{m}a and y=t^{n}b for some a,b ∈ C.
Now...
xy
= t^{m}a t^{n}b
= t^{m}t^{n}ab
=t^{n}t^{m}ba
= t^{n}b t^{m}a
= yx.
Notice we used the fact that a,b ∈ C to freely swap their positions with the other factors. 


Term
Classify all groups of order 45 

Definition
The Sylow subgroups are of orders 5 and 3^{2} = 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 H_{5} and H_{9}.
Since H_{5} and H_{9} are unique subgroups of their orders, they must be normal.
H_{5} is of prime order so it must be isomorphic to Z5 .
H_{9} is isomorphic to either Z_{3} x Z_{3} or Z_{9}.
The intersection H3 and H9 is trivial since only 1 divides both 5 and 9.
(H_{3})(H_{9}) is a group since H_{3} (alternatively, H_{9}) is normal.
Moreover, (H_{3})(H_{9}) must equal G since its order is a multiple of both 5 and 9.
Therefore G is the internal direct product of H_{5} and H_{9}.
Since there were two possible structures for H9 we have two possible structures for G:
Z_{5} x Z_{9} = Z_{45} or Z_{5} x Z_{5} X Z_{3} = Z_{3} x Z_{15}.
So G is either cyclic of order 45 or the product of two cyclic groups of orders 3 and 15. 


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

Definition
hk must be in H U K so in H or in K.
If hk in H then hk = h' > k = h^{1}h' So k in H.
Thus K is contained in H.
If hk in K then hk = k' > h = k^{1}k' So h in K.
Thus H is contained in K. 


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

Definition
If x is nilpotent, then 1 − x is a unit, because x^{n} = 0 entails
(1  x)(1 + x + x^{2} + ... + x^{n1}) = 1  x^{n} = 1
More generally,
the sum of a unit element and a nilpotent element
is a unit when they commute.



Term
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 zerodivisor.
(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. 

Definition
(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 zerodivisor.
(b) Suppose that x^{m} = 0 and y^{n} = 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_{ij}x^{i}y^{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^{m}x^{m} = 0 so rx is nilpotent.
(c) Suppose x^{m} = 0.
Then (1+x)(1−x+x^{2} − · · ·+ (−1)^{m−1}x^{m−1} ) = 1, so 1 +x is a unit.
If u is any unit and x is nilpotent, u + x = u(1 + u^{−1}x) is the product of two units (using that u^{−1}x is nilpotent by the above) and hence a unit. 


Term
Find conjugacy classes in S_{3} 

Definition
S_{3} = { () (12) (13) (23) (123) (132) }
a()a^{1} = () for any a in S_{3} 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 S_{3} and the conjugacy class
{(12), (13), (23)}
we can choose (12) = (12)(3) as our representative.
f_{1} is 1^{1}(1)! =1
f_{2} is 2^{1}(1)! = 4
f_{3} is 3^{0}(0)! = 1
So the size of our conjugacy class is 3!/2 = 3.



Term
Find conjugacy classes in S_{4} 

Definition
S_{4} = {
()
(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! / (f_{1}f_{2}...f_{n})
f_{j} is j^{aj}(a_{j})! with a_{j} = how many cycles of size j.
For the (ab)(cd) class
f_{1} is 1^{0}(0)! = 1
f_{2} is 2^{2}(2)! = 4
f_{3} is 3^{0}(0)! = 1
f_{4} is 4^{0}(0)! = 1
So the size of our conjugacy class is 4!/8 = 3.
For the (ab) class
f_{1} is 1^{2}(2)! = 2
f_{2} is 2^{1}(1)! = 2
f_{3} is 3^{0}(0)! = 1
f_{4} is 4^{0}(0)! = 1
So the size of our conjugacy class is 4!/4 = 6.
For the (abc)(d) class
f_{1} is 1^{1}(1)! = 1
f_{2} is 2^{0}(0)! = 1
f_{3} is 3^{1}(1)! = 3
f_{4} is 4^{0}(0)! = 1
So the size of our conjugacy class is 4!/3 = 8.
For the (abcd) class
f_{1} is 1^{0}(0)! = 1
f_{2} is 2^{0}(0)! = 1
f_{3} is 3^{0}(0)! = 1
f_{4} is 4^{1}(1)! = 4
So the size of our conjugacy class is 4!/4 = 6. 


Term
Prove Wilson's Theorem
(p1)! == 1 (mod p)
Also if (n1)!== 1 (mod n) then n is prime 

Definition
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, x^{2}  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 p1 to get 1 (mod p).
So (p1)! ≡ 1 (mod p).
For the converse? 


Term
State and prove Fermat's 2 square theorem. 

Definition
Fermat's [image] theorem, sometimes called Fermat's twosquare 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 (p1)/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 mth 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.



Term
Prove
If p = 4k+1
then p1 is a square (quadratic residue) (mod p).


Definition
Take (1)(4k)(2)(4k1)...(2k)(2k+1) = (p1)! = 1
(by Wilson's Theorem).
But
(1)(4k)(2)(4k1)...(2k)(2k+1)
= 1(1)2(2)...(2k)(2k) = ((2k)!)^{2}
(Note an even number of factors.)
So 1 = c^{2}
Where c = ((p1)/2)!.



Term
Factor ideal (2) in Z[sqrt(5)] 

Definition
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))(1sqrt(5)) but 2 doesn't divide either factor.
Suppose (2) = I^{2}.
Try I = (2, 1 + sqrt(5))


