Problem Solving - Groups

Problem Solving - Groups  

M.Velrajan

 

Set is the base of Abstract Mathematics. And Group is the Entrance of Abstract Algebra. So we start our discussion on Problem Solving - Pure Mathematics with elementary exercises on group theory using only the definition of a group given in Topics in Algebra, by I. N. Herstein, second edition.

Herstein before defining a group discusses the set A(S) of all bijections on a set S as a group. And he asks the readers to prove the following on A(S). 

1. If the set S has more than 2 elements then the group A(S) is non abelian.

Think how to prove it.

Note that A(S) is non abelian, if there exist at least two elements 

f, g ∈ A(S) such that fg ≠ gf. 

We are given S has more than 2 elements i.e. at least 3 elements. 

So using 3 elements of S, we have to get two elements f, g in A(S) 

such that fg ≠ gf.

3 elements of S — two non identity bijections f, g in A(S).

Why non-identity bijection ?  

Identity bijection commutes with all bijections.

A non-identity bijection should move at least two elements.

From 3 elements we can choose any two elements to move in a bijection and for the other bijection another two elements have to be chosen, so one of the already chosen elements has to be chosen again with the unchosen one.

Let a, b, c be three distinct elements of S. 

Define f, g : S → S by

f(a) = b, f(b) = a, 

f(x) = x for all x ∈ S - {a, b} and

g(b) = c, g(c) = b, 

g(x) = x for all x ∈ S - {b, c}. 

Then f, g ∈ A(S), 

f o g (a) = f(g(a)) = f(a) = b and  

g o f (a) = g(f(a)) = g(b) = c.

Hence f o g ≠ g o f and hence

A(S) is non abelian.

2. If the set S has n elements then the group A(S) has n! elements.

Suppose S = {x1, . . . , xn}.  

f ∈ A(S) is a map f : S → S which is one-one and onto. 

Since S is finite all one-one maps f : S → S are onto. 

And f is one-one iff f(x1),  . . . , f(xn) are distinct elements of S.

Any f ∈ A(S) is completely determined by f(x1),  . . . , f(xn) and 

f(x1),  . . . , f(xn) are distinct elements of S.

f(x1) can be any element of S, so f(x1) can be chosen in n ways.

f(x2) can be any element of S, but f(x2) ≠ f(x1), 

so f(x2) can be chosen in (n - 1) ways.

f(x3) can be any element of S, but f(x3) ≠ f(x1) and f(x3) ≠ f(x2), 

so f(x3) can be chosen in (n - 2) ways.

Proceeding thus we can see that 

f(xk) can be chosen in (n - k + 1) ways, 1 ≤ k ≤ n. 

Hence the number of elements in A(S) = n.(n - 1)(n - 2) . . . 2.1 = n! 

3. Herstein has given a proof for the following : 

If S = ℤ then the set G of all elements in A(S) which moves only a finite number of elements of S is a group.

And asks the reader to verify that G is an infinite and non abelian group.

Before verifying this we first observe that in the proof given nowhere it is used that the elements of the set S are integers. Hence the set G of all elements in A(S) which moves only a finite number of elements of S is a group for any infinite set S. 

So even from proof, by observation, we can get something more.

And also observe that G is a subgroup of A(S). Of course this can be observed only after knowing subgroups.

Now we verify the result. 

It can be verified by observations in the proof of 1.

In the proof of 1 observe that for any two distinct elements a, b ∈ S, 

there is an element f ∈ A(S) which moves only two elements of S 

i. e. there is an element f ∈ G. 

Since S is infinite there are infinite distinct pairs of elements of S 

and hence there are infinite elements in G. 

Also observe in the proof of 1 that 

for any three distinct elements a, b, c ∈ S, 

there are two elements f, g ∈ G such that f o g ≠ g o f 

and hence G is non abelian.

So, G is an infinite and non abelian group.

4. For any n > 2 construct a non-abelian group of order 2n.

(Hint : imitate the relation in S3.)

Herstein defines S3 as follows :

We observe that S3 = {e, 𝛗, 𝛙, 𝛙2, 𝛗.𝛙,  𝛙.𝛗}, 

where 𝛗2 = e, 𝛙3 = e, 𝛗.𝛙 = 𝛙-1.𝛗 = 𝛙2.𝛗.

Also 𝛗.𝛙2 = 𝛗.𝛙.𝛙 = 𝛙2.𝛗.𝛙 = 𝛙2.𝛙2.𝛗 = 𝛙4.𝛗 = 𝛙.𝛗.

Hence S3 = {e, 𝛗, 𝛙, 𝛙2, 𝛗.𝛙, 𝛗.𝛙2}, 

where 𝛗2 = e, 𝛙3 = e, 𝛗.𝛙 = 𝛙-1.𝛗 = 𝛙2.𝛗.

Think how to imitate these relations to get a 

non-abelian group of order 2n, for any n > 2.

Observe that S3 is a non-abelian group of order 6 = 2.3.

We have to construct a non-abelian group of order 2n. 

So replace 3 by n, 𝛗 by a symbol a and 𝛙 by b. 

Let G = {e, a, b, b2, b3, . . . , bn-1, ab, ab2, ab3, . . . , abn-1} 

be a set of formal symbols, 

where a2 = e, bn = e, ab = b-1a = bn-1a. 

i.e. G = {aibj / i = 0, 1; j = 0, 1, 2, . . . , (n - 1)}.

Then G is a non-abelian group of order 2n.

(verify! First express (aibj)(akbl) as arbs.)

In fact, this is also an exercise problem given in Topics in Algebra and it is called a dihedral group.

So, Herstein expects from the readers to observe and to think mathematically to get new results by themselves.

Observe - Think - Explore / Extend - Enjoy Learning.

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