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Abstract Algebra

Additional Mathematics Flashcards




Composition Series of G


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




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 (bm-1)b = b(bm-1) bm = e. So bm-1 is the inverse of b and it's in S since S is closed under ·.



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.




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.



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)}



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

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



∀ 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



In a finite group G, ∃N | ∀a∈G, aN = 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: aj = ak , for some j < k. Thus ak-j = e.



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.)


Sample Card Front
Sample Card Back

If ker(φ) = {0} then φ is an injection.
φ(a) = φ(b)

→ φ(a) - φ(b) = 0

→ φ(a - b) = 0

→ a - b = 0

→ a = b.


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 ><




Group homomorphism maps identity to identity


φ(a)φ(e)= φ(ae) = φ(a) = φ(a)e'


-->  φ(e) = e' by cancelation in G'.




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.






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



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 ∈ tmC and y ∈ tnC for some m,n.

i.e. x=tma and y=tnb for some a,b ∈ C.



= tma tnb

= tmtnab


= tnb tma

= yx.


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

Classify all groups of order 45

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.




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 + ... + xn-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 xm = 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 xm−1  0 by minimality of m. Thus x is a zero-divisor.

(b) Suppose that xm = 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 aijxiyj 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 = rmxm = 0 so rx is nilpotent.

(c) Suppose xm = 0.

Then (1+x)(1−x+x2 − · · ·+ (−1)m−1xm−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 Sn.  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 Sn ,

choose a permutation from the class

represented by its cyclic decompositon.

For each 

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

then fj  is jaj(aj)!

Now divide n! by the product of the fj.


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...fn)

 fj  is jaj(aj)!   with aj = 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 Zp is a field,

every nonzero element has a multiplicative inverse.

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

Also, because Zp 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


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 ωp, 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.



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).




= 1(-1)2(-2)...(2k)(-2k) = ((2k)!)2

(Note an even number of factors.)


So -1 = c2

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


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

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))



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