Unit - 2

Product of two subgroups

2.1 Product of two subgroups

Definition of subgroup-

Let G = {a, b, c,….} be a group with respect to o. Any non-empty subset G’ of G is called a subgroup of G if G’ is itself a group with respect to o.

Clearly G’ = {u}, where u is the identity element of G, and G itself are subgroups of any group G.

They will be called improper subgroups; other subgroups of G, if any, will be called proper. We note in passing that every subgroup of G contains u as its identity element.

Note-

1. A non-empty subset G0 of a group G is a subgroup of G if and only if (i) G’ is closed with respect to o, (ii) G’ contains the inverse of each of its elements.
2. Let a be an element of a group G. The set G’ = of all integral powers of a is a subgroup of G.
3. If S is any set of subgroups of G, the intersection of these subgroups is also a subgroup of G.

Product of two subgroups

Let H = and K  =   be subgroups of a group G and define the ‘‘product’’

Note- If H and K are invariant subgroups of a group G, so also is HK.

Key takeaways

1. Let G = {a, b, c,….} be a group with respect to o. Any non-empty subset G’ of G is called a subgroup of G if G’ is itself a group with respect to o.
2. If S is any set of subgroups of G, the intersection of these subgroups is also a subgroup of G.

2.2 Properties of cyclic groups

Cyclic group- A group G is said to be cyclic if G = [a] = for some .

For example: the additive group Z of integers and the additive groups Z/(n) of integers modulo n.

G is a cyclic group generated by a is written as- G = [a]

Note- A cyclic group is necessarily Abelian.

Example: The additive group J of integers is an infinite cyclic group, the integer, 1 being the generator.

Theorem: Every cyclic group is isomorphic to Z or to Z/(n) for some

Proof:

If G = [a] is an infinite cyclic group, consider the mapping given by .

It is clear that is a surjective homomorphism.

Moreover, for otherwise “a” would be of finite order.

Hence  is injective.

Therefore is an isomorphism.

Now let G = [a] is a cyclic group of finite order n.

Then G = and o(a) = n.

Consider the mapping given by.

is well defined and also injective, for let

Then

Here clearly is surjective.

Further,

Hence is an isomorphism.

Note- Any two cyclic groups of the same order are isomorphic.

Theorem: (Fundamental theorem of cyclic groups)- every subgroup of a cyclic group is cyclic.

Proof:

Suppose G = [a] be a cyclic group, and let H be a subgroup of G.

If H is a trivial subgroup, the result is obvious.

So let H is a proper subgroup of G.

If then . Hence, there is a least positive integer m such that .

We prove that H = [b].

Let by using division algorithm,

Then

Hence r = 0, therefore, , that proves H = [b].

Theorem: Let G be a finite cyclic group of order n, and let d be a positive divisor of n. Then G has exactly one subgroup of order d.

Proof:

The result holds trivially if d = 1 or d = n, so let 1 < d < n and put n/d = m. Let G = [a], then b = is of order d. Hence, [b] is a cyclic subgroup of order d.

To prove the uniqueness, let H be any subgroup of G of order d.

Then by the above theorem, H is generated by an element c = . Then . Hence n|sd, that is md|sd and so m|s.

Let mq = s, this yields . But since each of the subgroups [ and [ is of order d, [. Which proves H = [b].

Example: Suppose H = [a] and K = [b] be cyclic groups of order m and n respectively, such that (m, n) = 1. Then H is a cyclic group of order mn.

Sol:

Suppose order of a, b is d. Now implies d|mn.

Also (e, e)  = implies , so m|d and n|d. Therefore mn|d. Consequently, mn = d, since | H = mn, it follows that (a, b) generates the group H .

Key takeaways:

1. A group G is said to be cyclic if G = [a] = for some .
2. For example: the additive group Z of integers and the additive groups Z/(n) of integers modulo n.
3. A cyclic group is necessarily Abelian.
4. Every cyclic group is isomorphic to Z or to Z/(n) for some every subgroup of a cyclic group is cyclic.

2.3 Classification of subgroups of cyclic groups

Here we will consider the case for infinite and finite cyclic groups respectively.

1. Case-1 (infinite cyclic group)-

We are going to prove that every proper subgroup of an infinite cyclic group is itself an infinite cyclic group and accordingly isomorphic to the group itself.

Each member of a proper subgroup H of a cyclic group

G = {a}

Is some power of a.

Let m is the smallest non-negative integer such that

And is any arbitrary element of H.

There exists integers q and r such that

So that

Hence r = 0 and we have

So that k is a multiple of m.

Also, conversely, if k is any multiple of m, we have

Thus

Consisting, as it does, of the elements.

The equality of any two members of the set will imply that

Which, however not the case. Thus H is an infinite cyclic group.

2. Case-1 (finite cyclic group)-

Here we will prove that every subgroup of a finite cyclic group is cyclic and to every divisor l of the order of the group, there corresponds one and only one subgroup of that order

Let

Be a cyclic group of order n, so that it consists of the n elements

As in first case, we may show that if m be the smallest positive integer such that belongs to a given subgroup H of G.

Then H consists of the powers of so that

And is accordingly cyclic.

Let l be the order of H so that l is the least positive integer such that

And H consists of the l distinct elements

Of course, by Lagrange’s theorem, l is a divisor of n.

Thus every subgroup is cyclic.

Suppose now that l is any divisor of n and let n = lm.

Then

Is a subgroup of G of order l.

It can be obtained on taking the mth power of each element of G.

We have now to show that this is the only subgroup of order l.

This is a consequence of the fact that a subgroup of order l is obtained when we take the mth powers of the elements of G where m =n/l so that its relationship to G is given in terms of the order l only.

Theorem: every group of prime order is cyclic.

Assume that the order is greater than one, let a be any element other than the identity,

Then

Is a cyclic subgroup of the given group.

The order of H being a divisor of p, other than 1, is necessarily p and as such, we have

Theorem: Every finite group of composite order possesses proper subgroups.

Suppose G is a finite group of order lm where l and m are not 1.

If G is cyclic and a is any generator thereof, then

Is a proper subgroup of G, being of order m.

Now suppose G is not cyclic, then each independent generating system of G will contain atleast 2 elements.

Then the cyclic group generated by any member of an independent generating system is a proper subgroup of the given group.

Key takeaways

1. Every group of prime order is cyclic.
2. Every finite group of composite order possesses proper subgroups.

2.4 Cycle notation for permutations

Introduction

Cyclic notation was first introduced by the great French mathematician Cauchy in 1815.

Let us consider the permutation

We can present these as-

Although mathematically satisfactory, such diagrams are cumbersome.

Instead, we leave out the arrows and simply write = (1, 2) (3, 4, 6)(5).

Example, consider

In cycle notation, can be written (2, 3, 1, 5)(6, 4) or (4, 6)(3, 1, 5, 2), since both of these unambiguously specify the function . An expression of the form is called a cycle of length m.

A multiplication of cycles can be introduced by a cycle as a permutation that fixes any symbol not appearing in the cycle.

Thus, the cycle (4, 6) can be thought of as representing the permutation

In this manner, we can multiply cycles by thinking of them as permutations given in array form.

If array notations for and , respectively, are

Then, in cycle notation, = (12)(3)(45), (153)(24), and = (12)(3)(45)(153)(24).

To put in disjoint cycle form, observe that (24) fixes 1; (153) sends 1 to 5; (45) sends 5 to 4; and (3) and (12) both fix 4. So, sends 1 to 4. Continuing in this way we obtain 5 (14)(253).

One can convert back to array form without converting each cycle of into array form by simply observing that (14) means 1 goes to 4 and 4 goes to 1; (253) means 2 5, 5 3, 3 2.

2.5 Properties of permutations, even and odd permutations

Theorem- Every permutation of a finite set can be written as a cycle or as a product of disjoint cycles.

Proof:

Let be a permutation on A = {1, 2, . . . , n}. To write a in disjoint cycle form, we start by choosing any member of A, say , and let

…………

…………

…………

…………

And so on, until we arrive at for some m. We know that such an m exists because the sequence , . . . . Must be finite so there must eventually be a repetition, say for some i and j with i < j. Then

Where m = j - i. We express this relationship among

The three dots at the end indicate the possibility that we may not have exhausted the set A in this process. In such a case, we merely choose any element of A not appearing in the first cycle and proceed to create a new cycle as before.

That is, we let

And so on, until we reach for some k. This new cycle will have no elements in common with the previously constructed cycle. For, if so, then

For some i and j. But then

And therefore

This contradicts the way was chosen. Continuing this process until we run out of elements of A, our permutation will appear as

In this manner, we see that every permutation can be written as a product

Of disjoint cycles.

Theorem: If the pair of cycles and have no entries in common, then

Proof:

Suppose are permutations of the set 

Where the c’s are the members of S left fixed by both .

To prove we must show that for all x in S. If x is one of the a elements, say then

Since fixes all a elements.

For the same reason,

Hence, the functions of and agree on the a elements. A similar argument shows that and agree on the b elements as well.

Finally, suppose that x is a c element, say ci. Then, since both and

Fix c elements, we have

And

Which is the proof:

Note- (Order of a permutation)-  The order of a permutation of a finite set written in disjoint cycle form is the least common multiple of the lengths of the cycles.

Theorem: Every permutation in , n > 1, is a product of 2-cycles

Proof:

First, note that the identity can be expressed as (12)(12), and so it is a product of 2-cycles, By first theorem, we know that every permutation can be written in the form

A direct computation shows that this is the same as

Even and odd permutations:

Definition:

A permutation that can be expressed as a product of an even number of 2-cycles is called an even permutation. A permutation that can be expressed as a product of an odd number of 2-cycles is called an odd permutation.

Note- The set of even permutations in forms a subgroup of .

2.6 Alternating group

The group of even permutations of n symbols is denoted by and is called the alternating group of degree n.

Theorem: For n . 1, An has order n!/2.

Proof:

For each odd permutation a, the permutation (12) is even and (12) (12) when . Thus, there are at least as many even permutations as there are odd ones. On the other hand, for each even permutation a, the permutation (12) is odd and (12) (12) when . Thus, there are at least as many odd permutations as there are even ones. It follows that there are equal numbers of even and odd permutations. Since |Sn| = n!, we have |An| = n!/2.

The names for the symmetric group and the alternating group of degree n come from the study of polynomials over n variables. A symmetric polynomial in the variables x1, x2, . . . , xn is one that is unchanged under any transposition of two of the variables. An alternating polynomial is one that changes signs under any transposition of two of the variables.

Key takeaways

1. Every permutation of a finite set can be written as a cycle or as a product of disjoint cycles.
2. If the pair of cycles and have no entries in common, then
3. The order of a permutation of a finite set written in disjoint cycle form is the least common multiple of the lengths of the cycles.
4. The set of even permutations in forms a subgroup of .

References:

1. Contemporary abstract algebra, Joseph A. Gallian

2. Schaum’s abstract algebra

3. Basic abstract algebra by P.B. Bhattacharay, SK jain, S.R. Nagpaul

4. Modern abstract algebra by Shanti narayan, S.Chand & Co.