Unused Cycles

June 11, 2008

What is a mathematical group? (Part 3)

This will be my last post in this little series I’ve started on group theory. Here are parts one and two.

In part one, I defined the terminology needed to identify the axioms for a group. In part two, I gave examples of a group, and began a short taste of what group theory is all about (doing things abstractly). In this part, I’ll giving an only slightly more involved introduction to proofs in group theory, as well as introduce some other so-called “algebraic structures” other than groups.

Previously, I showed that the identity element is unique within a group. Is the same true for an inverse of an element within a group? That is, given an element a within a group, is there only one element a^{-1} such that a^{-1}a=aa^{-1}=e? The claim is:

Theorem: An element a within a group G has a unique inverse.
Proof: Again the argument is by contradiction. Suppose that there are two elements that have this property, b and c. Then


We can define the power of an element in a group, a^n to be \underbrace{aaa\ldots a}_{n\text{ times}} just as expected. We can also define it inductively so that a^n=a^{n-1}a for positive n.

Note that there is nothing in the definition of the group G that states that it is necessary that ab=ba for an a,b\in G. This is, in general, not true. Recall that non-singular matrices under matrix multiplication are a group. But matrix multiplication is not always commutative. A group for which every element commutes with every other element is called abelian. Just a small theorem involving abelian groups is the following:

Theorem: If a^2=e for every element a\in G, then G is abelian.
Proof: We can take two elements a,b so that ab\in G since we’re inside a group. Then by assumption (ab)^2=e. But we also have e=ee=a^2b^2 so that (ab)(ab)=(aa)(bb). Right multiplying by b^{-1} and left multiplying by a^{-1} gives ba=ab, the desired equality.

The theory of groups is much more involved and complicated than just simple theorems that show a certain property such as uniqueness of an element or that one property implies another. For further reading on the topic of groups, try reading about cosets, Lagrange’s Theorem, permutations, and isomorphism of groups. You can also try reading the online textbook that we used for my course in abstract algebra here, though be warned it goes more in depth than the topics presented here – it’s definitely for somebody who is not just mildly interested in the subject.

In closing, I’ll introduce other common algebraic structures without doing anything with them.

Definition: A commutative ring is a set R with two binary operations, + and \ast that has the following properties for every a,b,c\in R:

  1. Additive Commutativity: a+b=b+a
  2. Additive Associativity: (a+b)+c=a+(b+c)
  3. Additive Identity: There exists an element, 0, such that a+0=0+a=a
  4. Additive Inverses: There exists an element -a such that a+(-a)=(-a)+a=0
  5. Multiplicative Commutativity: ab=ba
  6. Multiplicative Associativity: a(bc)=(ab)c
  7. Multiplicative Identity: There exists an element 1 such that 1a=a1=a
  8. Distributivity: a(b+c)=ab+ac

Notice that this definition says that the set is an abelian group under addition, but it is important to note that it is not a multiplicative group (there is no guarantee of a multiplicative inverse). Examples of rings include the non-negative integers with addition and multiplication, rational numbers with addition and multiplication, and the real numbers with addition and multiplication. The set of n\times n real matrices with matrix multiplication and addition do not form a commutative ring since they are not commutative under multiplication. They do, however, form a “non-commutative” ring.

The next definition “fixes” multiplication in a ring so that it is (almost) a multiplicative group.

Definition: A field is a commutative ring where every non-zero element a\in G has a multiplicative inverse.

Of the previous rings identified, only the rational and real numbers are fields.

The last definition I’m going to give will be a familiar one to most physicists (as they’ve almost surely taken a course in linear algebra).

Definition: A vector space over a field F is a set V of elements called vectors together with a binary operation + on V and a function \ast: F\times V\rightarrow V called scalar multiplication (the asterisks is again usually dropped). Given a,b\in F and x,y,z\in V, the following properties hold:

  1. Additive Associativity: (x+y)+z=x+(y+z)
  2. Additive Commutativity: x+y=y+x
  3. Additive Identity: There exists a vector, called 0, such that x+0=x
  4. Additive Inverse: There exists a vector -x such that x+(-x)=(-x)+x=0
  5. Vector Distributivity: a(x+y)=ax+ay
  6. Scalar Distributivity: (a+b)x=ax+bx
  7. Multiplicative Associativity: (ab)x=a(bx)
  8. Multiplicative Identity: Given the multiplicative identity in F, 1x=x

The reader familiar with linear algebra knows that F=\mathbb{C} and V=\mathbb{C}^n is a vector space. Other examples include F=\mathbb{R} and V as the set of all polynomials of degree less than or equal to n.

That’s it for this introduction to abstract algebra. Give me your thoughts if you found it interesting!

(Part 1)(Part 2)



  1. […] 1)(Part 2)(Part 3) Possibly related posts: (automatically generated)What is a mathematical group? (Part […]

    Pingback by What is a mathematical group? (Part 2) « Unused Cycles — June 11, 2008 @ 6:25 pm

  2. […] What is a mathematical group? (Part 3) In this part, I’ll giving an only slightly more involved introduction to proofs in group theory, as well as introduce some other so-called “algebraic structures” other than groups. Previously, I showed that the identity element is … […]

    Pingback by » What is a mathematical group? (Part 3) 3 Some: What The World Is Saying About 3 Some — June 11, 2008 @ 6:57 pm

  3. Sorry for a stupid question:

    Where do those names come from? Are there any intuitive reasons for using them? For example, why is the object as defined called ring but not something else? Any relation to the ordinary meaning of the word?

    Comment by Ben — August 25, 2008 @ 7:19 pm

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