<h1>Introduction to Abstract Algebra: Part 3</h1>
<hr>
<p>
Last year I took the classes Abstract Algebra I and II from the great Joseph Gallian who literally wrote <a href="https://www.google.com/search?tbm=isch&q=abstract%20algebra%20textbook&tbs=imgo:1">the book on abstract algebra</a>.  I loved the classes and learned a lot.  This is why I decided to write this series of posts.
<br><br>

<a href="https://steemit.com/mathematics/@jackeown/introduction-to-abstract-algebra-part-1">Part One - Introduction to Groups</a>
<a href="https://steemit.com/mathematics/@jackeown/introduction-to-abstract-algebra-part-2-cyclic-groups-and-subgroups">Part Two - Cyclic Groups and Subgroups</a>

</p>
<hr>
So in <a href="https://steemit.com/mathematics/@jackeown/introduction-to-abstract-algebra-part-2-cyclic-groups-and-subgroups">Part Two</a> 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 <b><i>Symmetric Group</i></b> which has to do with permutations.


<hr><h2>Proof that Identities are Unique:</h2>
(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!
<hr>

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!

<hr><h2>Fundamental Theorems of Cyclic Groups</h2>
I realize now that my last post may have glossed over these...a lot...and I'm probably still going to here...
<hr>
The following are important things to note about groups that I forgot to mention:
<ul>
<li>The order of an element divides the order of the group.</li>
<li>Every subgroup of a cyclic group is cyclic</li>
<li>The order of every subgroup divides the order of the original group. <i><b>(Lagrange's theorem)</b></i></li>
<li>For finite abelian groups, there is a unique subgroup with order d for every divisor, d, of |G|.</li>
</ul>

Also...a <i><b>"coset"</b></i> 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}.
<p>
1*H is the set formed by combining 1 and each element in H with the group operation. <br>
1*H = {1*0,1*3,1*6,1*9,1*12} = {1,4,7,10,13}<br>
2*H = {2*0,2*3,2*6,2*9,2*12} = {2,5,8,11,14}<br>
and
3*H = {3*0,3*3,3*6,3*9,3*12} = {3,6,9,12,0} = H<br>
</p>

<h6>Armed with this knowledge...if you want proofs of Lagrange's theorem and others above...check out these links:</h6>
<a href="https://en.wikipedia.org/wiki/Coset">Cosets</a>
<a href="https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)">Lagrange's Theorem</a>
(plus most of these proofs can be found by simple googling...then the answers are mostly found on the Math stackexchange ;))


<hr><h2>Introduction to Symmetric Groups:</h2><hr>

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:
<pre><code>Two-Line Notation
(1 2 3 4 5)
(4 3 2 5 1) 
</code></pre>
----------------------
<pre><code>One-Line Notation
(4 3 2 5 1)
</code></pre>
----------------------

<pre><code>Cycle Notation
(1 4 5)(2 3)
</code></pre>
<br>

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 <b><i>where the displaced people go</i></b>.
For a good visual explanation of how this works, check out this video on it: 
https://www.youtube.com/watch?v=MllHjO5Pntk

<hr><h2>So how is this a group?</h2><hr>

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)

<h4>Watch this video to understand how to multiply permutations: </h4>
https://www.youtube.com/watch?v=gGir2mwNrbo


This is a group, once again, because it satisfies all four group axioms:
<ul>
<li><b>Closure</b>: if a,b are in G, then a*b is in G </li>
<li><b>Associativity</b>: (a*b)*c = a*(b*c)</li>
<li><b>Identity</b>: G must have one element, e, where e*b = b for all b in G <br>(i.e. a "do nothing" element)</li>
<li><b>Inverses</b>: For every element, a, in G there must be a b in G such that a*b = e</li>
</ul>


<h6>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.
</h6>

Thanks,
JacKeown