Introduction to Groups and Permutation Groups

Property of Amit Amola. Should be used only for reference and with consent.
We all know in full depth what groups are and what are their properties.
A group is a nonempty set together with a binary operation in such a way
that there is an identity element in that group, there is associativity in its
elements, every element has an inverse, and any pair of element can be
obtained without going outside the set (closure). With this definition now
let us try to understand Groups in more detailed manner with a basic
example.
Take this hexagonal “Ok” sign for an example:-
Let’s see that in how many ways we can reposition this hexagonal “OK”
sign-
By 60° rotation either towards left or right we can find six other positions.
The next 60° rotation would give back to us the initial position itself. Now
we shall check the other positions which we can get by flipping or
reflecting the hexagon through its side’s midpoints and through its vertices.
Abstract Algebra
Assignment
R0 R60 R120 R180 R240 R300
V3V2V1 M1 M2 M3

These are the only 12 positions which we can find or form by repositioning
or reflecting the hexagon. Now we can see that every motion which we
perform will give any of these 12 outcomes only. For example let’s rotate
the initial position by 120° and then reflect it through first of the three
main vertices.
We can clearly see that the final image is V3 only. Which means V1R120 = V2
So these 12 motions along with their compositions form a mathematical
system called the Dihedral Group. In this case as the number of elements or
total number of motions are 12 so the order of this Dihedral Group is 12 i.e.
this is the Dihedral Group of order 12. And it is denoted as D6. Now we shall
understand what groups exactly are with the help of the example we have
taken above.
For this we use a much simpler approach by constructing an operation
table or which we refer to as Cayley Table to see and observe the different
composition of motions which we can find.
R0 R60 R120 R180 R240 R300 V1 V2 V3 M1 M2 M3
R0 R0 R60 R120 R180 R240 R300 V1 V2 V3 M1 M2 M3
R60 R60 R120 R180 R240 R300 R0 M1 M2 M3 V2 V3 V1
R120 R120 R180 R240 R300 R0 R60 V2 V3 V1 M2 M3 M1
R180 R180 R240 R300 R0 R60 R120 M2 M3 M1 V3 V1 V2
R240 R240 R300 R0 R60 R120 R180 V3 V1 V2 M3 M1 M2
R300 R300 R0 R60 R120 R180 R240 M3 M1 M2 V1 V2 V3
V1 V1 M1 V2 M2 V3 M3 R120 R240 R0 R180 R300 R60
V2 V2 M2 V3 M3 V1 M1 R240 R0 R120 R300 R60 R180
V3 V3 M3 V1 M1 V2 M2 R0 R120 R240 R60 R180 R300
M1 M1 V2 M2 V3 M3 V1 R180 R300 R60 R240 R0 R120
M2 M2 V3 M3 V1 M1 V2 R300 R60 R180 R0 R120 R240
M3 M3 V1 M1 V2 M2 V3 R60 R180 R300 R120 R240 R0
R120 V1

These are the inferences we can take out from this Cayley table:-
 We can see in this table that all the outcomes are no different than the
already known motions. This means that any motion which we carry out
will result in these 12 outcomes only; means if P and Q are in D6 then PQ
is also in D6. This property is referred to as closure, and this is one of the
necessary requirement for a particular set to be defined as a group.
 We also see that in this set of motions i.e. D6 there exist an identity
element (here R0) such that if there is any other element let’s say P of D6,
then R0P = PR0 =P. So there must be an identity element in a set for it
being called as a group.
 Now we also see that for each element P in D6 there is exactly one
element Q in D6 such that PQ = QP = R0. Here P and Q are known as
inverse of each other. For example here in our table inverse of R240 is
R120 and vice versa and like R180 and V2 are their own inverses. Again this
is an important condition for a set to be referred as group.
 One more condition which is compulsory for a set to be referred as
group is associativity in its elements; i.e. (pq)r = p(qr) where p,q and r є
D6. There are 123= 1728 possible choices of p,q and r in D6 in which we
have to check associativity. Here it is not humanly possible but as we
know that the given motions are functions itself and the operation here
is of function composition. And since function composition is always
associative, so we take it as obvious that all the elements of our dihedral
group undergo associativity.
 One another keen observation here shows that every element of D6
appears exactly once in each row and column. This feature is something
that all groups must possess and moreover we should keep this fact in
mind while making Cayley table.
 And the last observation is about that if the group is commutative or not;
i.e. if it’s any choice of group of elements let’s say P and Q are such that
PQ = QP then the group is commutative which is referred to as Abelian
groups. And if they aren’t commutative then they are just called Non-
abelian groups.

So we just saw the three necessary conditions which a non-empty set must
possess to be referred as a group beside it being closure; which namely are
associativity in its elements, presence of an identity element in that set &
availability of inverses of every element of the set.
Now let’s discuss Dihedral Groups in detail.
Dihedral Groups
In simple way a Dihedral Group is a group whose elements are symmetries
of a regular polygon and these symmetries are obtained by rotations and
reflections of the polygon from its vertices and mid-points of its sides.
Dihedral Groups are considered to be the simplest example of finite groups.
We can construct the similar symmetrical figures for other polygons too
with sides ≥3. And these groups are denoted as Dn and is called dihedral
group of order 2n, where 2n are the total number of elements in that group.
If the polygon has n sides, then it has 2n symmetries: n rotational
symmetries and n reflective symmetries. Let’s take another example which
I’ve taken from Wikipedia. Here is an Octagonal “Stop” sign which is
dihedral group of order (2x8) =16. So here are the 16 symmetrical figures
of Octagonal “Stop” sign.
The first row shows the effect of the
eight rotations, and the second row
shows the effect of the eight reflections.
Similarly a snowflake has D6 dihedral
symmetry, the same as a regular
hexagon.

Dihedral Groups are everywhere around us. Take example of various
company’s logos.
Like logo of Chrysler’s logo has D5 symmetry.
Mercedes-Benz logo has D3 as symmetry group.

Either take human made Rose Windows in a church as an example:-
Like this Rose Window shows Group of Symmetry of order 18, i.e. D9.
Or take for example Crop Circles which are presumed to be made by alien.

Well that was all about Dihedral Groups. Now before we talk about
Permutation Groups, let’s give a quick look on some other important terms
in a simplified manner.
Order of a Group- The number of elements in a group are called that group’s
order. And is denoted as |G| if we have to denote the order of the group G.
For example |D6|= 12.
Order of an element- The order of an element say m in a group say M is the
smallest positive integer n s.t. mn = e. If no such n exists then we say m has
infinite order. The order of m is denoted as |m|. For example order of R60 in
D6 is 6; i.e. | R60|= 6.
Subgroups- If there’s a subset H of a group G which is itself a group under
the operation of G, we say H is a subgroup of G. We denote it as H<G. There
are various methods to test whether the subset of a group taken under
consideration is a subgroup or not. Here I am just mentioning what those
tests are:-
 One-Step Subgroup Test
 Two-Step Subgroup Test
 Finite Subgroup Test
Cyclic Groups- A group say F is called cyclic if there is an element ‘a’ in F s.t.
F= {an | n є Z}. And such an element ‘a’ is called a generator of F. We also
write F as F= ⟨a⟩.
Now knowing these important terms we shall proceed to Permutation
Groups.
Permutation Groups
These are one of the important class of groups. One reason for their
importance is that most of the groups can be represented as a group of
permutations on a suitable set. (Exception include Quaternion Groups)
Now let’s get to know what Permutation Groups are-
A permutation group is a finite group let’s say G whose elements are
permutations of a given set and whose group operation is composition of
permutations in G. There is one-to-one mapping from G onto G. Well to
explain it better let’s see a very interesting example.

Let’s consider a set X containing 3 objects, say a triangle, a circle and a
square. A permutation of X = {△, O, □} might send for example
△ △, O □, □ O
and we observe that what just did is exactly to define a bijection on the set
X, namely a map σ: X X defined as
σ (△) = △, σ ( O) = □, σ (□) = O
Picture representation:-
X= { , , }
Now defining an arbitrary bijection:-
Now such function can be denoted through this notation
which is known as two-row form:-
( )
And if we denote these symbols as numbers then we can
also write this as:
(
𝟏 𝟐 𝟑
𝟏 𝟑 𝟐
)

So just above we saw one form of set X in which we can write its elements.
There are 5 other ways. For example let’s take set Y= {1, 2, 3} and let’s
discuss its 6 six permutation forms. I am using two row form to represent
these forms:
a= (
1 2 3
1 2 3
) b= (
1 2 3
2 3 1
) c= (
1 2 3
3 1 2
)
d= (
1 2 3
2 1 3
) e= (
1 2 3
3 2 1
) f= (
1 2 3
1 3 2
)
So these are the six permutation forms which we can make out from set Y.
And these are the six elements of its permutation group. Such kind of
groups are also known as Symmetry Groups and denoted as Sn.
Here we can also see that (b.f)≠(f.b), so that S3 is not an abelian group.
Now before I explain about this group’s identity element and the inverse, I
ought to introduce about composition of the elements of a permutation
group. In composing permutations we always follow the same convention
we use in composing any other mappings: read from right to left. For
example let,
σ= (
1 2 3
3 1 5
4 5
4 2
) λ= (
1 2 3
4 3 1
4 5
2 5
)
σλ = (
1 2 3
3 1 5
4 5
4 2
) (
1 2 3
4 3 1
4 5
2 5
) = (
1 2 3
4 5 3
4 5
1 2
)
We see that finally on right we have 4 below 1, it’s so because
(σλ)(1)= σ(λ)(1) = σ(4)= 4, so σλ sends 1 to 4. And other elements are
obtained in similar manner.
Now as we know how composition in elements of permutation group
works, so let’s talk about identity element and inverse of S3 group. An
identity element should be of a kind such that if any element of S3 say ‘m’ is
operated with it then it gives ‘m’ only as a result.
So here a= (
1 2 3
1 2 3
) is the identity element.
Let’s say we take element ‘b’, so we can see that a.b=b.a=a.

Now let’s talk about inverse of these elements. Let’s find inverse of element
c.
c= (
1 2 3
3 1 2
) c-1= (
1 2 3
2 3 1
)
Thus, the inverse of an element is obtained by reading from the bottom
entry to the top entry rather than from top to bottom: if 1 appears beneath
3 in c then 3 appears beneath 1 in c-1.
So this was about symmetric group S3. Let’s talk little more about
Symmetric groups in general.
Symmetric Groups (Sn)
Let A= {1, 2, 3, …, n}. The set of all permutations of A is called the
symmetric group of degree n and is denoted by Sn. Elements of Sn have the
form
є= [
1 2
є(1) є(2)
… … … …
𝑛
є(n)]
It is easy to compute the order of Sn. There are n choices of є(1). Once є(1)
has been determined, there are (n-1) possibilities for є(2)[since є is one-to-
one, we must have є(1)≠ є(2)]. After choosing є(2), there are exactly (n-2)
possibilities for є(3). Continuing along in this fashion, we see that Sn must
have n(n-1)…3.2.1= n! elements.
Now, since S1 = {(1)} then S1 with respect to composition is commutative.
Similarly, since
[
1 2
1 2
] [
1 2
2 1
]= [
1 2
2 1
] [
1 2
1 2
]
then S2 = {[
1 2
1 2
] [
1 2
2 1
]}is also Abelian.
Unfortunately, this is not true anymore for |S| > 2.
We will now prove that Sn is not abelian when n≥3.

Theorem- Sn is non-Abelian for n≥3.
Proof: All that we need to do here is to find two permutations σ and λ in Sn
with n≥3 such that (σ ○ λ) ≠ (λ ○ σ). Indeed, consider the permutations
σ= (
1 2
1 3
3 4 5
2 4 5
… … …
𝑛
𝑛
) and λ= (
1 2
3 2
3 4 5
1 4 5
… … …
𝑛
𝑛
)
we know that of course such two elements will always exist in a symmetric
group where n≥3. And moreover we can see that
σ ○ λ= (
1 2
2 3
3 4 5
1 4 5
… … …
𝑛
𝑛
) and λ ○ σ= (
1 2
3 1
3 4 5
2 4 5
… … …
𝑛
𝑛
)
i.e. σ ○ λ≠ λ ○ σ
Hence for n≥3, Sn is always non-abelian.
Cycle Notation for Permutations
There’s one more way we write or denote the elements of permutation. It is
called as Cycle Notation. The cycle notation for permutations can be
thought as a condensed way to write permutations. Here is how it works.
Let α є Sn be the permutation
and α(a1) = a2, α(a2) = a3, · · · , α(ak) = a1
i.e. α follows the circle pattern
Such a permutation is called a cycle of length k or simply a k-cycle. We will
write
α= (a1a2a3……ak)
The cycle notation is read from left to right, it says α takes a1 into a2, a2 into
a3, etc., and finally ak, the last symbol, into a1, the first symbol.
a1
ak
a2
a3
..
.
..
...
..
.

Let’s see how this cycle notation works with an example:
Let say we have an element as
α= (
1 2
1 6
3 4 5
3 7 5
6
4
7
2
)
Some points to get acquainted with:
 So this element α can also be written in cycle notation as
α= (1)(2647)(3)(5) or just by a 4-cycle notation (2647)
It is customary to omit the terms whose image is same as their own.
Moreover this notation can be used in many ways.
α= (2647)
= (6472)
= (4726)
= (7264)
 This thing above shows us that a k-cycle can be written in k different
ways, since
(a1a2a3……ak)= (a2a3 a4……ak a1) = … = (aka1 a2……ak-1)
 We can easily find the inverse of a cycle. Since α(ak)= ak+1 implies
α-1(ak+1)= ak , we only need to reverse the order of the cyclic pattern. For
example,
(2647)-1 = (7462)
Now let’s get acquainted with how the multiplication of cycles are done.
Multiplication of cycles- Multiplication of cycles is performed by applying
the right permutation first.
Consider the product in α= (12)(3)(45) and β= (153)(24) and α β=
(12)(3)(45) (153)(24)

Now from reading right to left,
1 1 5 4 4 4
So 1 4.
Now
4 2 2 2 2 1
So 4 1.
Now
2 4 4 5 5 5
So 2 5.
Now
5 5 3 3 3 3
So 5 3
Now
3 3 1 1 1 2
So 3 2. So we can see that this multiplication can be represented as
αβ= (14)(253)
i.e. αβ= (12)(3)(45) (153)(24)= (14)(253)
We just saw that in the original multiplication term there were some cycles
with same element. But in the final result there is no such thing. So those
two cycles are called disjoint cycles.
There are two theorems which are associated with disjoint cycles. Let’s see
what they say.
Theorem: If α and β are disjoint cycles then αβ = βα.
Proof-
Indeed, since the cycles α and β are disjoint, each element moved by α is
fixed by β and vice versa. Let α = (a1a2 · · · as) and β = (b1b2 · · · bt) where
{a1, a2, · · · , as} ∩ {b1, b2, · · · , bt} = ∅.
(i) Let 1 ≤ k ≤ s. Then
(αβ)(ak) = α(β(ak)) = α(ak) = ak+1
and
(βα)(ak) = β(α(ak)) = β(ak) = ak+1
(ii) Let 1 ≤ k ≤ t. Then
(αβ)(bk) = α(β(bk)) = α(bk+1) = bk+1
and
(βα)(bk) = β(α(bk)) = β(bk) = bk +1

(iii) Let 1 ≤ m ≤ n and m ∉{a1, a, … as, b1, b2, … , bt}. Then
(αβ)(m) = α(β(m)) = α(m) = m
and
(βα)(m) = β(α(m)) = β(m) = m
It follows from (i), (ii), and (iii) that αβ = βα.
So that was theorem which says that if α and β are disjoint cycles then αβ =
βα. Now we have one more theorem which tells that every permutation of a
finite set can be written as a cycle or product of disjoint cycle.
Theorem: Every permutation of a finite set can be written as a
cycle or as a product of disjoint cycles.
Proof-
Let α be a permutation on A = {1, 2, … ,n}. To write α in disjoint cycle form,
we start by choosing any member of A, say a1 and let
a2= α(a1), a3= α(α(a1)) = α2(a1),
and so on, until we arrive at a1 = αm(a1) for some m. We know such an m
exists because the sequence a1, α(a1), α2(a1), . . . must be finite; so there
must eventually be a repetition, say αi(a1) = αj(a1) for some i and j with
i < j. Then a1= αm(a1), where m = (j – i). We express this relationship
among a1,a2, . . . . . ., am as
α= (a1,a2, . . . . . ., am) . . .
The ellipsis 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 an
element b1 of A not appearing in the first cycle and proceed to create a new
cycle as before. That is, we let b2 = α(b1), b3= α2(b1), and so on, until we
reach b1 = αk(b1) for some k. This new cycle will have no elements in
common with the previously constructed cycle. For, if so, then
αi(a1) = αj(b1) for some i and j. But then αi-j(a1)= b1 and therefore b1= at for
some t. This contradicts the way b1 was chosen. Continuing this process
until we run out of elements of A, our permutation will appear as
α= (a1,a2, . . . , am)( b1,b2, . . . , bk) . . . (c1,c2, . . . , cs).
In this way, we see that every permutation can be written as a product of
disjoint cycles.

Now there’s one more theorem which tells us about order of a permutation.
Theorem: The order of a permutation of a finite set written in
disjoint cycle form is the LCM of the lengths of the cycles.
Proof-
We can easily see that a cycle of length n has order n.
Now let’s suppose that α and β are two disjoint cycles of lengths m and n
respectively, and let k be the least common multiple of m and n. Now we
have read a theorem earlier (not proved in this article) which is stated as-
Let G be a group and let ‘a’ be an element of order n in G. If ak = e,
then n divides k.
So from this theorem it follows that both αk and βk are the identity
permutation ε and, since α and β commute, (αβ)k = αkβk is also the identity.
Thus, we know by the corollary to the above theorem (ak = e implies that
|a| divides k) that the order of αβ(let’s say it t) must divide k. But then
(αβ)t= αtβt = ε, so that αt = β-t. However, it is clear that if α and β have no
common symbol, the same is true for αt and β-t, since raising a cycle to a
power does not introduce new symbols. But if αt and β-t are equal and have
no common symbols, they must both be the identity, because every symbol
in αt is fixed by β-t and vice-versa (a symbol not appearing in a permutation
is fixed by the permutation). It follows, then, that both m and n must divide
t. This means that k, the least common multiple of m and n, divides t also.
This shows k = t.
Thus far, we have proved that the theorem is true in the cases where
the permutation is a single cycle or a product of two disjoint cycles. The
general case involving more than two cycles can be handled in similar
fashion.

Now there’s an another theorem which is like this-
Theorem: Every permutation in Sn, n > 1, is a product of 2-cycles.
Proof:
Note that the identity can be written as (12)(12).
A k-cycle (a1a2 . . . ak) can be written as
(a1a2 . . . ak) = (a1ak)(a1ak−1) · · · (a1a3)(a1a2).
Since any permutation can be written as a product of disjoint cycles, we can
decompose any permutation into a product of transpositions by
decomposing each disjoint cycle in the product.
Examples:
(12345) = (15)(14)(13)(12) or (54) (52) (21) (25) (23) (13).
(1632)(457) = (47) (45) (12) (13) (16).
One more interesting theorem tell us that:
Theorem: If a permutation α can be expressed as a product of even
number of 2-cycles, then every decomposition of α into a product
of 2-cycles must have an even number of 2-cycles. In symbols, if
α = β1β2 . . . βr and α = γ1γ2 . . . γs,
where the β’s and the γ’s are 2-cycles, then r and s are both even or
odd.
Proof-
Observe that β1β2 . . . βr = γ1γ2 . . . γs implies
ε = γ1γ2 . . . γs βr
-1 . . . β2-1β1-1
= γ1γ2 . . . γs βr . . . β2β1,
since a 2-cycle is its own inverse. Thus, the theorem above guarantees that
s+r is even. It follows that r and s are both even or both odd.
And this leads to these definitions;
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.

These above two theorems show that every permutation can be without
any doubt classified as either even or odd, but not both.
This observation has a very big significance. In next theorem i’ll show that
why we needed the above observations.
Theorem: Even permutations form a group
or
The set of even permutations in Sn forms a subgroup of Sn.
Proof-
These set of even permutations occur so often that we refer to them by a
special name as Alternating Group of Degree n or An.
Now applying the finite subgroup test to An:
We apply the Finite Subgroup Test to An:
As per finite subgroup test we need to prove that a-1∈ An whenever a∈An.
We can see e = (12)(12) is an even permutation of Sn. So e ∈ An.
And we know any 2-cycle permutation element γ ∈ An will have γ-1 in An
only.
Now we also see that whether or not it satisfies closure property. Let
α,β ∈ An. Then α is a product of an even number, say 2k, of 2-cycles and β is
a product of an even number, say 2l, of 2-cycles.
Then αβ is a product of an even number, 2(k+l), of 2-cycles i.e. even
permutation. Hence αβ∈ An.
Hence this proves that the set of even permutations in Sn forms a subgroup
of Sn.
Now the next result shows that exactly half of the elements of Sn(n>1) are
even permutations.

Theorem: For n>1, An has order n!/2.
Proof-
We know that cycle (12) denotes identity. Now for each odd permutation σ,
the permutation (12)σ is even. Thus there are at least as many even
permutations as there are odd ones. On the other hand, for each even
permutation ϕ the permutation (12) ϕ is odd. So, there are at least as many
odd permutations as there are even ones. It
follows that there are an equal number of even
and odd permutations.
Since |Sn|= n!., thus we have |An|=n!/2.
Example of use of Alternating Groups:
The 12 rotations of a Tetrahedron can be
conveniently described with the elements of A4.

Finally a quick look at the history and development of permutation groups.
History of Permutation Groups
The study of groups originally grew out of an
understanding of permutation
groups. Permutations had themselves been
intensively studied by Lagrange in 1770 in his
work on the algebraic solutions of polynomial
equations.
This subject flourished and by the mid-19th
century a well-developed theory of permutation
groups existed, codified by Camille Jordan in his
book of 1870. Jordan's book was, in turn, based on
the papers that were left by Évariste Galois in 1832.
When Cayley introduced the concept of an
abstract group, it was not immediately clear
whether or not this was a larger collection of
objects than the known permutation groups
(which had a definition different from the
modern one). Cayley went on to prove that the
two concepts were equivalent in Cayley's
theorem.
Joseph Louis Lagrange
Camille Jordan
Évariste Galois
Arthur Cayley

Another classical text containing several chapters on permutation groups
is Burnside's Theory of Groups of Finite Order of 1911. The first half of the
twentieth century was a fallow period in the study of group theory in
general, but interest in permutation groups was revived in the 1950s by
H. Wielandt whose German lecture notes were reprinted as Finite
Permutation Groups in 1964.
So that’s all I covered in this article. We talked about Groups by giving
example of Dihedral Groups and then discussed them in detail. Then in
between we also talked about some important terms and then we moved
ahead to Permutation Groups where we first introduced about Permutation
Groups and then also introduced Symmetry Groups. We also saw many
theorems and observations and made inferences out of them. So that’s
what we learnt about in this article.
The End
William_Burnside
Helmut Wielandt

Reference:
Books:
 Contemporary Abstract Algebra (4th ed.) by Joseph A. Gallian
 The Mathematics of Identification Numbers by Joseph A. Gallian
 A First Course in Abstract Algebra with Applications (3rd ed.) by
Rotman, Joseph J
 Abstract Algebra (3rd ed.) by Dummit, David S., Foote, Richard M.
 Symmetry Groups and Their Applications by Miller, Willard Jr.
 Symmetry by H. Weyl
Websites:
 Mathigon- http://world.mathigon.org/
 Wikipedia- http://en.wikipedia.org/
 Priceton University - www.princeton.edu/
 http://math.berkeley.edu/
 http://mathworld.wolfram.com/
 www.google.com

Introduction to Groups and Permutation Groups

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