Introduction to Abstract Algebra: Part 3

Last year I took the classes Abstract Algebra I and II from the great Joseph Gallian who literally wrote the book on abstract algebra. I loved the classes and learned a lot. This is why I decided to write this series of posts.

Part One - Introduction to Groups
Part Two - Cyclic Groups and Subgroups

So in Part Two I sort of glossed over a bunch of proofs and details that I should have included, but I didn't want the post to get too long. Because of this, I've decided to make Part 3 be a review of what we've learned so far (and should have...) as well as an introduction to another very important group known as the Symmetric Group which has to do with permutations.

Proof that Identities are Unique:

(i.e. there is exactly one element,e, in any group such that e*a=a=a*e for all a in the group.) I forgot to show this simple proof, but I think it helps understand and it's just kinda neat!

For the sake of argument, assume that a group, G, has two identities, e1, and e2.
This means that (e1*a = e2*a = a) for all a in G.
Now multiply everything (on the right) by a^(-1)
This gives us e1*a*a^(-1) = e2*a*a^(-1) and we can group it as follows because of associativity!
e1*(a*a^(-1)) = e2*(a*a^(-1)) which just becomes
e1 = e2

Therefore our assumption at the beginning that we could have two identities is false!

Fundamental Theorems of Cyclic Groups

I realize now that my last post may have glossed over these...a lot...and I'm probably still going to here...

The following are important things to note about groups that I forgot to mention:

The order of an element divides the order of the group.
Every subgroup of a cyclic group is cyclic
The order of every subgroup divides the order of the original group. (Lagrange's theorem)
For finite abelian groups, there is a unique subgroup with order d for every divisor, d, of |G|.

Also...a "coset" is a set of elements you get from taking every element of a subgroup combined with a given element from the group.

For instance let G = Z15 = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14}.
Let H be the subgroup H = {0,3,6,9,12}.

1*H is the set formed by combining 1 and each element in H with the group operation.
1*H = {1*0,1*3,1*6,1*9,1*12} = {1,4,7,10,13}
2*H = {2*0,2*3,2*6,2*9,2*12} = {2,5,8,11,14}
and 3*H = {3*0,3*3,3*6,3*9,3*12} = {3,6,9,12,0} = H

Armed with this knowledge...if you want proofs of Lagrange's theorem and others above...check out these links:

Cosets
Lagrange's Theorem
(plus most of these proofs can be found by simple googling...then the answers are mostly found on the Math stackexchange ;))

Introduction to Symmetric Groups:

Symmetric groups are pretty cool.
They model permutations (or different ways to arrange stuff).
There are a few different notations for them.
There's 2-line notation, 1-line notation, and disjoint cycle notation.

The following all mean the same thing:

Two-Line Notation
(1 2 3 4 5)
(4 3 2 5 1)

----------------------

One-Line Notation
(4 3 2 5 1)

----------------------

Cycle Notation
(1 4 5)(2 3)

Imagine that we are arranging 5 people in a row for a photo.
There are 5 people as well as 5 spots to place them in.
We can label the people and the positions.

The above permuation (which originally is probably easiest to understand in 2-line notation) means the following:
Person 1 goes in Position 4.
Person 2 goes in Position 3.
Person 3 goes in Position 2.
Person 4 goes in Position 5.
Person 5 goes in Position 1.

Since the top row of this is largely redundant, we can just get rid of it to yield the 1-line notation version.

Alternatively, disjoint cycle notation makes a lot less sense, but it is convenient when we come to see permutations as group elements. With disjoint cycle notation we look at it this way:
Person 1 goes to Position 4.
Person 4 goes to Position 5.
Person 5 goes to Position 1.
---- end cycle ------------------------
Person 2 goes to Position 3.
Person 3 goes to Position 2.
---- end cycle ------------------------

In other words, we don't just go down the line, but instead we look at where the displaced people go.
For a good visual explanation of how this works, check out this video on it:

So how is this a group?

The preceding paragraphs detailed how to represent a single permutation.
The set of ALL permutations on n elements form a group known as Sn ("the Symmetric Group on n elements").
As such, the preceding permutations were all elements of S5.

In order to assess the "groupiness" of this supposed group, we must first say what "*" means in this context.
It makes sense that p*q would mean permutation p being applied to the identity and subsequentially applying q to the resulting state.

So in 1-line notation:
(2 3 1)*(3 2 1) = (1 3 2)

In cycle notation: (with cycle notation you multiply right to left.)
(1 2 3)*(1 3)(2) = (1)(2 3)

Watch this video to understand how to multiply permutations:

This is a group, once again, because it satisfies all four group axioms:

Closure: if a,b are in G, then a*b is in G
Associativity: (a*b)*c = a*(b*c)
Identity: G must have one element, e, where e*b = b for all b in G
(i.e. a "do nothing" element)
Inverses: For every element, a, in G there must be a b in G such that a*b = e

Try it out yourself to convince yourself...or write up a nice proof for the comments!
If you have any questions or suggestions for future topics, feel free to comment on that as well.

Thanks,
JacKeown

Introduction to Abstract Algebra: Part 3 - Review and The Symmetric Group