Abstract Algebra 2: Groups

Well my last article was terrible. It was hastily thrown together just to have something, but really only one thing was important. Recall that I defined an operation \ast on elements in some set S to be associative if for any 3 elements in some set S, call the elements a,b, and c, the following is always true.

(a\ast b ) \ast c = a\ast (b  \ast c)

The operation is the thing which is associative. Consider the set \mathbb{Z} = \{\ldots ,-2,-1,0,1,2,\ldots\} with the operation of regular addition. That is, for any two elements in \mathbb{Z} we consider addition between them to be the sum.

This operation on this set is associative. For example: (4+3) +2 = 4 + (3 + 2)

And this is true for any 3 numbers, always. The key is that we are only looking at one operation. Remember that when you start mixing addition with multiplication, suddenly this doesn’t work. It has to be the same operation between the 3 elements for it to remain true.

That is the obvious example that most people are familiar with, but we want to look at generalities not at specific cases. Associativity is seemingly arbitrary, we just like it to be true so we suppose it is true and then see what happens. There are two more “good” properties that we like.

The first is the identity element. The identity element is the element that when you put it with other elements, doesn’t change anything. In the above example, 0 is the identity. When you add 0 to something, nothing changes.

More formally: For a set S, we say that an element “e” is an identity element if for any other element “a” in the set S, we have that
a \ast e = e \ast a = a
This is exactly what 0 does in the above example, but in a completely general sense. Instead of looking at the integers we are talking about any arbitrary set and instead of 0, just some element that has the same property.

The last “good” concept that we like to have is the existence of an inverse element for any given element. An inverse element is something that you combine with another element to end up with the identity. In the integer example, if I am given the element 5, then the inverse of 5 is -5 because putting them together with the operation of addition yields 0, the identity element of the example. In general, given any element “a” in the integers, -a is the inverse of a. So this example is “nice” because every single element has an inverse.

Formally we write: For a set S, we say that an element (-a) is the inverse of a if a \ast -a = -a \ast a = e where e is the identity element of S.

In conclusion we have three nice properties; associativity, existence of an identity, and an inverse element for any element in the set. These three simple and seemingly routine properties turn out to automatically generate a bunch of other much more interesting things, many of which are not instantly recognizable. When a set S has these three properties we say that S is a group.

One example is that it turns out that when you have a group, the identity is unique. That is, there is only one identity for any given group. For example, 0 is the only identity for the integers under addition. No other element can have the same property. This is a provable thing, but it is not something that was assumed. That is the basis of math, assume as few things as possible and prove a bunch of things based only on the assumptions.

Lastly, try to think of other examples of groups. What about the same set with the operation of multiplication instead of addition?

Well it is still true that (a \cdot b) \cdot c = a \cdot (b \cdot c), but 0 is no longer the identity. If you multiply 0 with anything, you get 0, not the number you started with.

Instead, the number 1 satisfies the condition to be the identity element. 1 \cdot a = a \cdot 1= a is always true for any element a in the set of integers. However, this is not a group because we lack inverses. Under multiplication the inverse of 5 is \frac{1}{5} but \frac{1}{5} is not an integer. In fact, for most elements a, we have that \frac{1}{a} is the multiplicative inverse, but \frac{1}{a} is not in the set of integers.

Therefore, the integers under the operation of multiplication do not form a group.

I guess I forgot to mention this concept of closure, where we are only looking at the set. \frac{1}{a} is outside of the set we are talking about so it is as if it doesn’t exist. So although \frac{1}{a} has an inverse in some bigger set, it does not have one in the set we are talking about, which is why the set we are looking at is not a group even though there might be a bigger group which does have these elements.

Advertisements
This entry was posted in Uncategorized. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s