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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.

If G is a group and

∀ a,b ∈ G,  (a⋅b)² = a²⋅b²

then G is Abelian.

Proof

Let a and b be elements of G and suppose 

(a⋅b)² = a²⋅b² ,

which may be written

 a⋅b⋅a⋅b = a⋅a⋅b⋅b.

We have

 a⋅b   = e⋅a⋅b⋅e   =   a^-1⋅a⋅a⋅b⋅b⋅b^-1

=  a^-1⋅a⋅b⋅a⋅b⋅b^-1   =  e⋅b⋅a⋅e  =  b⋅a.

Therefore G is Abelian.

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.)

If G is a finite group and p is a prime number dividing the order of G (the number of elements in G), then G contains an element of order p. That is, there is x in G so that p is the lowest non-zero number with x^p = e, where e is the identity element.

Wiki

Theorem: Let G be a finite group and p be a prime. If p divides the order of G, then G has an element of order p.

We first prove the special case that where G is abelian, and then the general case; both proofs are by induction on n = |G|, and have as starting case n = p which is trivial because any non-identity element now has order p. Suppose first that G is abelian. Take any non-identity element a, and let H be the cyclic group it generates. If p divides |H|, then a^|H|/p is an element of order p. If pdoes not divide |H|, then it divides the order [G:H] of the quotient group G/H, which therefore contains an element of order p by the inductive hypothesis. That element is a class xH for some x in G, and if m is the order of x in G, then x^m = e in G gives (xH)^m = eH in G/H, so p divides m; as before x^m/p is now an element of order p in G, completing the proof for the abelian case.

In the general case, let Z be the center of G, which is an abelian subgroup. If p divides |Z|, then Z contains an element of order pby the case of abelian groups, and this element works for G as well. So we may assume that p does not divide the order of Z; since it does divide |G|, the class equation shows that there is at least one conjugacy class of a non-central element a whose size is not divisible by p. But that size is [G : C_G(a)], so p divides the order of the centralizer C_G(a) of a in G, which is a proper subgroup because a is not central. This subgroup contains an element of order p by the inductive hypothesis, and we are done.

 Wiki

Prove

If G/Z(G) cyclic then G is abelian

Suppose G/Z(G) is cyclic.

Then by definition: ∃τ∈G/Z(G):G/Z(G)=⟨τ⟩

Since  τ is a coset by Z(G):

∃t∈G:τ=tZ(G)

Thus each coset of Z(G) in G is equal to

(tZ(G))ⁱ=tⁱZ(G) for some i∈Z. 

Now let x,y∈G.

Suppose x∈t^mZ(G),y∈tⁿZ(G).

Then x=t^mz₁,y=tⁿz₂ for some z₁,z₂∈Z(G).

Thus:  

xy 

= t^mz₁tⁿz₂

= t^mtⁿz₁z₂

= t^m+nz₁z₂

= t^n+mz₂z₁

= tⁿt^mz₂z₁

= tⁿz₂t^mz₁

= yx

Prove

For n ≥ 3 Center of S_n is trivial.

Let us choose an arbitrary π∈Sn:π≠e,π(i)=j,i≠j.

Then π(j)≠j since permutations are injective.

Since n≥3,

we can find k≠j,k≠π(j) and ρ∈Sn

which interchanges j and k and fixes everything else.

Let π(j)=m. Then m≠j,m≠k so ρ fixes m.

Then ρπ(j)=ρ(m)=m=π(j) since ρ fixes m.

Now k=ρ(j) by definition of ρ.

So π(k)=πρ(j).

But π(j)≠π(k) since permutations are injective.

Thus ρπ(j)≠πρ(j).

So arbitrary π≠e is not in the center

since there exists a ρ which it does not commute with.

Thus only e is in the center, which, by definition, is trivial.

Prove 

If C is the commutator subgroup of G then

G/C is abelian.

Wiki

Let xC,yC∈G/CxC,yC∈G/C 

then xyx−1y−1C=C

since xyx−1y−1

is a commutator hence belongs to C.

But then xyC=yxCso xCand yCcommute in G/CG/C.

This can be done for any elements, so G/CG/C is abelian.

Also should show C is normal.

Prove

Only one additive identity  

Cancelation:  ac = bc →a = b

Unique solution for ax = b 

Inverse of inverse is element 

Prove

Only one additive identity  e' = ee' = e

Cancelation:  ac=bc → ac(c')=bc(c') → ae=be→ a=b

Unique solution for ax = b → x = a'b

Inverse of inverse is element 

-(-a) = 0 + -(-a) = a + -a +  -(-a) = a + 0 = a

<g> is a subgroup of G so |<g>| must be odd (Lagrange).

Therefore g^2k-1 = e for some k.

Thus  (g^k)² = g^2k =  g(g^2k-1) = g(e) = g.

Hence g is the square of g^k.

Let G be the set of integers.

Determine whether G is a group under subtraction.

(G,-) is not a group under subtraction. The following axioms fail:

Associative law: 

(0 - 1) - 2 = -1 - 2 = -3 ≠ 1 = 0 - -1 = 0 - (1 - 2).

Therefore - is not associative within G.

Existence of an identity:

Suppose e is the identity of (G,-).

Then e - 1 = 1 - e, so e = 1.

Similarly e - 2 = 2 - e, so e = 2.

Since 1 ≠ 2, this is impossible.

Therefore G cannot have an identity. 

Existence of inverses.

Since (G,-) does not have an identity, 

it's not possible for any element to have an inverse in (G,-)

Let G be the set of all rational numbers with odd denominators.

Prove that (G,⋅) is a group where ⋅

is the usual addition

for rational numbers: +.

Proof:

Let a/b and c/d be elements of G,

So a,b,c and d are integers

with b and d odd.

a/b + c/d  = (ad + bc) / ad

 is an element of G since both

ad + bc  is an integer

and ad is odd, since

a product of odd integers

is always odd.

Hence (G,⋅) is closed.

Since addition is associative

for the rational numbers in general,

it's also associative for elements of G.

Therefore ⋅ is associative.

0 = 0/1 is an identity in (G,⋅)

since 0+x=x+0

for any rational number.

Let a/b be in G, where b is odd.

Then -(a/b) = -a/b ∈ G.

Therefore every element of (G,⋅)

has an inverse.

We've shown that (G,⋅) is a group.

Show that if every element of G

is it's own inverse then G is Abelian.

PROOF 

Suppose every element of G

is it's own inverse.

Let a and b be two elements of G.

We have

a⋅b = e⋅a⋅b⋅e = (b⋅b)⋅a⋅b⋅(a⋅a)

= b⋅(b⋅a)⋅(b⋅a)⋅a = b⋅e⋅a = b⋅a.

Therefore G is Abelian.

Let G be the set of all real 2-by-2 diagonal matrices of the form

where a is nonzero.

Prove that G is an Abelian group under matrix multiplication.

Let G be the set of all real 2-by-2 

matrices of the form

where a is nonzero. 

Prove that (G,⋅) is an Abelian group, where

 the ⋅ operator is

matrix multiplication among elements of G. 

PROOF 

Since 2-by-2 matrix multiplication is associative

 in general, it is associative within G.

Therefore ⋅ is associative.

Let A =

and B = 

be elements of G.  A⋅B  =

is an element of G (since ab is nonzero).  

Thus G is closed under the ⋅ operation.

Moreover, B⋅A  =

=

= A⋅B. 

Thus ⋅  is commutative.  

Let I=

Prove that if G is an Abelian group

then (a⋅b)ⁿ = aⁿ⋅bⁿ , ∀n ∈ ℤ.

PROOF

Suppose G is Abelian.

First we prove by induction that 

∀a,b ∈G, (a⋅b)ⁿ = aⁿ⋅bⁿ,  ∀n ≥ 0 .

Basis step

(a⋅b)⁰ =  e = e⋅e = a⁰⋅b⁰ .

So the property holds when n = 0.

Inductive step

Suppose  (a⋅b)^k = a^k⋅b^k for a given k≥ 0.

Then

(a⋅b)^k+1 = (a⋅b)^k ⋅(a⋅b)

= (a^k⋅b^k)⋅(a⋅b)

= (a^k⋅a)⋅(b^k⋅b)   (since ⋅ is commutative)

= a^k+1⋅b^k+1.

We've shown

∀a,b ∈G, (a⋅b)ⁿ = aⁿ⋅bⁿ,  ∀n ≥ 0.

Now suppose n < 0.

Then (a⋅b)ⁿ = (a⋅b)^--n

= ((a⋅b)^-1)^-n

= (b^-1⋅a^-1)^-n (by lemma 2.3.1d)

= (a^-1⋅b^-1)^-n  (since ⋅ is commutative)

=     (a^-1)^-n⋅(b^-1)^-n (by the above, since -n > 0.)

= a^--n⋅b^--n 

= aⁿ⋅bⁿ.

Thus  for any a,b in G,

(a⋅b)ⁿ = aⁿ⋅bⁿ, ∀n ∈ ℤ.

²Write out all the right cosets of H in G where...

a) G = <a> of order 10 and H = <a²> .

b) G as in part a and H = <a⁵>.

c) G = A(S) where S = {x₁, x₂, x₃} and H = σ∈G such that σ(x₁) = x1.

6) Write out all the right cosets of H

in G where...

a) G = <a> of order 10 and H = <a²> .

b) G as in part a and H = <a⁵>.

c) G = A(S) where S = {x₁, x₂, x₃}

and H = σ∈G such that σ(x₁) = x1.

__________________________________

a) H = <a²> = {e, a², a⁴, a⁶, a⁸}.

Ha = {e⋅a, a²⋅a, , a⁴⋅a, , a⁶⋅a, , a⁸⋅a}

= {a, a³, a⁵, a⁷, a⁹}.

Since H ∪ Ha = G, 

and right cosets of a subgroup

partition the group, H and Ha 

is a complete listing of right cosets of H.

b) H = <a⁵> ={e, a⁵}.

Ha = {e⋅a, a⁵⋅a} = {a, a⁶}.

Ha² = {e⋅a² , a⁵⋅a² } = {a² , a⁷}.

Ha³ = {e⋅a³ , a⁵⋅a³ } = {a³ , a⁸}.

Ha⁴ = {e⋅a⁴ , a⁵⋅a⁴ } = {a⁴ , a⁹}.

Since H ∪ Ha ∪ Ha² ∪ Ha³ ∪ Ha⁴   = G

 H , Ha , Ha² , Ha³ and Ha⁴   

is a complete listing of right cosets of H.

c) The members of A({x₁,x₂,x₃})

expressed in cycle notation are

(), (x₁ x₂), (x₁ x₃), (x₂ x₃), (x₁ x₂ x₃}

and (x₁ x₃ x₂).

Only () and (x₂,x₃} fix x₁.

Thus H = {(), (x₂ x₃)},

H(x₁ x₂)={()(x₁ x₂), (x₂ x₃)(x₁ x₂)}

= {(x₁ x₂) (x₁ x₃ x₂)}

and

H(x₁ x₃)={()(x₁ x₃), (x₂ x₃)(x₁ x₃)}

= {(x₁ x₃) (x₁ x₂ x₃)}.

Since H ∪ H(x₁ x₂) ∪ H(x₁ x₃)  = G,

  H, H(x₁ x₂) and H(x₁ x₃)

is a complete listing of the right cosets of H.

7c) Recall G = A(S) where S = {x₁, x₂, x₃}

and H = σ∈G such that σ(x₁) = x1.

As in Problem 6c, H= {(), {x2 x3}}.

Now

(x1 x2)H = {(x1 x2)(), (x1 x2)(x2 x3)}

= {(x1 x2) X1 x2 x2)} and

(x₁ x₃)H={(x₁ x₃)(), (x₁ x₃)(x₂ x₃)}

= {(x₁ x₃) (x₁ x₃ x₂)}.

Since H ∪ (x₁ x₂)H ∪ (x₁ x₃)H  = G,

H, (x₁ x₂)H and (x₁ x₃)H

is a complete listing of left cosets of G.

S1) If H and K are subgroups of a group G

such that ○(H) and ○(K) are relatively prime,

prove that H ∩ K ꞊ {e}.

Since H and K are subgroups of G,

so is H ∩ K.  

(Herstein Section 2.4 exercise 1.)

Moreover H ∩ K is a subgroup of H

(since H ∩ K ⊆ H)

and  H ∩ K is a subgroup of K

(since H ∩ K ⊆ K). 

H and K have finite orders so

by the Lagrange theorem,

o(H ∩ K) divides o(H) 

and 

o(H ∩ K) divides o(K).

Since gcd(o(H), o(K)) = 1,

o(H ∩ K) = 1.

Thus H ∩ K  = {e}.

S1) If H and K are subgroups of a group G

such that ○(H) and ○(K) are relatively prime,

prove that H ∩ K ꞊ {e}.

Since H and K are subgroups of G,

so is H ∩ K.  

(Herstein Section 2.4 exercise 1.)

Moreover H ∩ K is a subgroup of H

(since H ∩ K ⊆ H)

and  H ∩ K is a subgroup of K

(since H ∩ K ⊆ K). 

H and K have finite orders so

by the Lagrange theorem,

o(H ∩ K) divides o(H) 

and 

o(H ∩ K) divides o(K).

Since gcd(o(H), o(K)) = 1,

o(H ∩ K) = 1.

Thus H ∩ K  = {e}.

S2. Let G be a group of order 4.

Prove that either G is cyclic or x^2 ꞊ e

for every x in G.

Conclude that G must be Abelian.

Since G if finite,

by Corollary 1 to Lagrange's theorem,

the order of each element divides

the order of G. Thus for any x in G, o(x) ∈ {1,2,4}.

If o(x) = 4. then G is cyclic.

Otherwise

o(x) = 1  (so x = e)

or o(x) = 2.

In either case x⋅x = e.

Now, if G is cyclic, G is Abelian.

(For any generator a of G,

a^ka^j = a^k+j = a^ja^k.)

On the other hand, if x⋅x = e, ∀x∈G,

then given a, b ∈ G we have

a⋅b = e⋅(a⋅b) =  (b⋅a)⋅(b⋅a)⋅(a⋅b)

= b⋅a⋅b⋅a⋅a⋅b = b⋅a⋅b⋅e⋅b

= b⋅a⋅b⋅b= b⋅a⋅e =  b⋅a.

Thus G is Abelian.

Jf a cyclic subgroup T of G is normal in G,

then show that every subgroup of T is normal in G.

Let T be a cyclic subgroup of G,

with T = <a>, for some a ∈ G.

Suppose T is normal in G,

so gag^-1 = a^j ∈ T,

for some integer j. 

Let U be subgroup of T,

so  U = <a^k >,

for some integer k.

Then for any (a^k)^m = a^km ∈ U

and  for any g in G

g(a^km)g^-1 = (gag^-1)^km 

= (a^j)^km = a^jkm = (a^k)^jm ∈ U.

So U is normal in G.