Unit – 2
Product of two subgroups
Q1) What are subgroups and product of subgroups?
A1)
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.
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.
Q2) Define cyclic group.
A2)
A group G is said to be cyclic if G = [a] = 
for some 
.
Foe 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]
Example: The additive group J of integers is an infinite cyclic group, the integer, 1 being the generator.
Q3) Prove that every cyclic group is isomorphic to Z or to Z/(n) for some 
A3)
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.
Q4) State and prove fundamental theorem of cyclic groups.
A4)
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].
Q5) 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.
A5)
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 
.
Q6) Show that every group of prime order is cyclic.
A6)
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
Q7) Show that Every finite group of composite order possesses proper subgroups.
A7)
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.
Q8) What do you understand by the cyclic notation for permutations.
A8)
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.
Q9) Prove that every permutation of a finite set can be written as a cycle or as a product of disjoint cycles.
A9)
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.
Q10) Prove that Every permutation in 
, n > 1, is a product of 2-cycles.
A10)
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
Q11) Define even and odd permutations.
A11)
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 
.
