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Modern Algebra: Groups, Rings, Modules, Fields

Sam Macdonald

Section 1.1 Group Basics

“Most great learning happens in groups.”
―Ken Robinson

Subsection Binary Operations and the Definition of a Group

We begin, as all great volumes of mathematics are like to do, somewhere in the middle.
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Though everything that you need from the beginning can be found in Appendix A.

Definition 1.4. Binary Operation.

A binary operation on a set \(S\) is a Function
\begin{equation*} -*-:S\times S\to S,\text{ given by } (x,y)\mapsto x * y. \end{equation*}
One of the nice things about binary operations is that they have properties.

Definition 1.5. Binary Properties.

Let \(-*-\) be a binary operation on a set \(S\text{.}\)
  1. An operation satisfies the associative property if for all \(x,y,z\in S\) we have \((x*y)*z=x*(y*z)\text{.}\)
  2. An operation satisfies the identity property if there exists \(e\in G\) such that \(e*g=g*e=g\) for all \(g\in G\text{.}\) Such an element \(e\) is called an identity element.
  3. An operation satisfies the inverse property if for each \(x\in S\text{,}\) there is an element \(y\in S\) such that \(x * y=e=y * x\text{,}\) where \(e\) is an identity element of \(S\text{.}\) Such an element \(y\) is called an inverse of the element \(x\text{.}\)

Definition 1.6. Group.

A group is a pair \((G, *)\) where \(G\) is a set and \(*\) is a binary operation on \(G\) satisfying the associative, identity, and inverse properties.

Mark 1.7.

We say that “\(S\) is closed under the operation \(*\)”, when we want to emphasize that for any \(x,y\in S\) the result of the operation, \(x*y\text{,}\) is an element of \(S\text{.}\) However note that closure is really part of the definition of a binary operation on a set, and it is implicitly assumed whenever we consider such an operation.
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This is the one and only "Mark". All following "Marks" will be "Remarks".

Subsection First Properties of Groups

Proof.

  1. Suppose there exist two elements \(e\) and \(e'\) such that \(ex = x = xe\) and \(e'x = x = xe'\) for all \(x\text{.}\) As \(e\) is an identity element, we have \(e'e=e'\text{.}\) Similarly, as \(e'\) is an identity element, we have \(e'e=e\text{.}\) Combining these two equations, we see
    \begin{equation*} e = ee' = e'. \end{equation*}
    Thus \(e=e'\text{,}\) making the identity element unique.
  2. Suppose an element \(x\) has two inverses: \(y\) and \(z\text{.}\) Thus \(yx = xy = e\) and \(zx = xz = e\text{.}\) Observe
    \begin{align*} z &= ez & & \knowl{./knowl/xref/identity-property.html}{\text{\(e\) is the identity element}}\\ &= (yx)z & & \text{substituting } yx = e \\ &= y(xz) & & \knowl{./knowl/xref/associative-property.html}{\text{Associative Property}}\\ &= ye & & \text{substituting } xz = e\\ &= y & & \knowl{./knowl/xref/identity-property.html}{\text{\(e\) is the identity element}} \end{align*}
    Thus \(y=z\text{,}\) making inverses unique.
  3. Suppose \(xy = xz\) for some \(x,y,z\in G\text{.}\) By the inverse property, \(x\) has an inverse, \(x\inv\in G\) such that \(xx\inv=e=x\inv x\text{.}\) Observe
    \begin{align*} xy &= xz \\ x\inv(xy) &= x\inv(xz) && \text{multiply on the left by }x\inv \\ (x\inv x)y &= (x\inv x)z && \knowl{./knowl/xref/associative-property.html}{\text{Associative Property}} \\ (e)y &= (e)z && \text{substituting } x\inv x = e\\ y &= z && \knowl{./knowl/xref/identity-property.html}{\text{\(e\) is the identity element}}. \end{align*}

Subsection Abelian Groups

Additionally, a binary operation can be commutative.

Definition 1.9. Commutative Property.

An operation satisfies the commutative property if \(x * y = y * x\) for all \(x,y \in S\text{.}\)

Definition 1.10. Abelian Group.

An abelian group is a group that also satisfies commutative property.

Convention 1.11. Additive vs. Multiplicative Notation.

When working with arbitrary groups, multiplicative notation is often used as the operation. Thus, groups are commonly written as \((G,\cdot)\) and referred to as multiplicative groups. However, when working with abelian groups, it is customary to use additive notation, writing the group as \((G,+)\text{.}\) Often the letter \(A\) is used in place of \(G\) when referring to abelian groups.

Subsection First Examples of Groups

Example 1.12. Group Examples.

  1. The trivial group is the group with a single element \(\{e\}\text{.}\)
  2. \((\Z,+), (\Q,+), (\R,+),\) and \((\C,+)\) each form an abelian group, where \(+\) denotes traditional addition. The axioms of arithmetic guarantee the validity of the group axioms as well as the commutativity of the group operation. Thus all four groups are abelian.
  3. For any positive integer \(n\text{,}\) let
    \begin{equation*} \operatorname{GL}_n(\C)=\{ \text{invertible } n \times n \;\text{matrices with entries in}\; \C\}. \end{equation*}
    Then \(\operatorname{GL}_n(\C)\) is a non-abelian group under matrix multiplication known as the general linear group.
  4. For any \(n\in\N\text{,}\) let \(\Z/n\) denote the The Integers Modulo \(n\). Then \((\Z/n,+)\) forms an abelian group where \(+\) denotes addition modulo \(n\text{.}\)
  5. The quaternion group \(Q_8\) is a set with \(8\) elements
    \begin{equation*} Q_8=\{ 1, -1, i, -i, j, -j, k, -k \} \end{equation*}
    satisfying the following relations: \(1\) is the identity element,
    \begin{equation*} i^2 = -1, j^2 = -1, k^2 =-1, ij = k, jk = i, ki = j, \text{and} \end{equation*}
    \begin{equation*} (-1)i = -i, (-1)j = -j, (-1)k = -k, (-1)(-1) = 1. \end{equation*}

Remark 1.13. Specific General Linear Groups.

In general, \(\GL_n(F)\) is a group for any field \(F\) (whatever those are).
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To jump ahead and discover what those are, see: Definition 8.1.

Convention 1.14. Identity Theft.

In familiar groups of numbers, “\(1\)” is used for the identity element of a multiplicative group (and likewise “\(0\)” in additive groups). Often, “\(1\)” is used for the identity element \(e\) in arbitrary groups, though we will not do so here.

Exercise 1.15. Not Quite Groups.

  1. The natural numbers \(\N\) do not form a group with respect to addition.
  2. The rational numbers \(\Q\) do not form a group with respect to multiplication.

Subsection Groups of Units

The following example illustrates a few important examples of something called a Group of Units, which is defined explicitely in terms of rings (whatever those are) in the aptly named Ring Theory.

Example 1.16. Groups of Units.

Each of the following subsets form an abelian group under multiplication.
  1. The subset \(\Z^\times=\{-1,1\}\) of \(\Z\text{.}\)
  2. \(\Q^\times=\Q\setminus\{0\}\text{,}\) \(\R^\times=\R\setminus\{0\}\text{,}\) and \(\C^\times=\C\setminus\{0\}\text{.}\)
  3. For each \(n\in\N\text{,}\) the subset \(\Z/n^\times=\{k\in\Z/n|\gcd(k,n)=1\}\) of \(\Z/n\text{.}\)

Convention 1.17.

Some texts use the notation “\(^*\)” to denote groups of units. For example, \(\Q^\times\) would be written \(\Q^*\text{.}\)

Remark 1.18.

\(\Z/n^\times\) will be integral to our constructions of semidirect products in Section 7.2 and our classifications of certain groups up to isomorphism (whatever that means). Though not strictly required for this text, knowing the structure of \(\Z/n^\times\) for various \(n\) can prove helpful in many qualifying exam problems.

Exercise 1.19. \(\Z/p^\times\).

Prove that \(\Z/p^\times=\mathbb{Z}/p\setminus\{0\}\text{.}\)

Subsection Other Initial Examples and Properties

Exercise 1.20. Some Extra Groups.

  1. Let \((G,*)\) be a multiplicative group, and define the opposite group, \(G^{op}\text{,}\) to be the set \(G\) equipped with the operation \(*^{op}\text{,}\) where \(x*^{op}y=y*x\text{.}\) Then \(G^{op}\) is a group.
     4 
    Opposite groups are used in category theory to describe something called “duality” between algebraic structures.
  2. Let \(\E\) denote the set of even integers and \(\O\) denote the set of odd integers. Then \((\E,+)\) is a group, whereas \((\O,+)\) is not.
Solution.
    • Associativity of \(G^{op}\).
      Let \(x, y, z\) be any elements \(G^{op}\text{.}\) Observe
      \begin{align*} (x*^{op}y)*^{op}z &= (y*x)*^{op}z & & \text{substituting }x*^{op}y=y*x\\ &= z*(y*x) & & \text{substituting again} \\ &= (z*y)*x & & \knowl{./knowl/xref/associative-property.html}{\text{Associative Property of \(G\)}}\\ &= (y*^{op}z)*x & & \text{substituting }z*y = y*^{op}z\\ &= x*^{op}(y*^{op}z) & & \text{substituting again} \end{align*}
      Thus, associativity is satisfied in \(G^{op}\text{.}\)
    • Identity Element of \(G^{op}\).
      As \(G\) is a group, it contains an identity element, \(e\text{.}\) I claim that \(e\) is also the identity element of \(G^{op}\text{.}\) Let \(x\in G^{op}\text{.}\)
      \begin{align*} e*^{op}x &= x*e & & \text{substituting }e*^{op}x=x*e\\ &= x & & \knowl{./knowl/xref/identity-property.html}{\text{\(e\) is the identity element of \(G\)}} \\ &= e*x & & \knowl{./knowl/xref/identity-property.html}{\text{\(e\) is the identity element of \(G\)}}\\ &= x*^{op}e& & \text{substituting }x*^{op}e=e*x \end{align*}
      Thus \(e*^{op}x = x*^{op}e = x\) for all \(x\in G^{op}\) Therefore, \(e\) is the identity element of \(G^{op}\text{.}\)
    • Inverses in \(G^{op}\).
      Let \(x\) be an element in \(G^{op}\text{.}\) Since \(x\) is also an element of the group \(G\text{,}\) by the inverse property there exists an inverse element \(x\inv\) in \(G\) such that \(x*x\inv = x\inv*x = e\text{,}\) where \(e\) is the identity element of \(G\text{.}\) Consider \(x\inv*^{op}x\)
      \begin{align*} x\inv*^{op}x &= x*x\inv & & \text{substituting }x\inv*^{op}x=x*x\inv\\ &= e & & \text{substituting }x\inv*x=e \end{align*}
      and \(x*^{op}x\inv\)
      \begin{align*} x*^{op}x\inv &= x\inv*x & & \text{substituting }x*^{op}x\inv=x\inv*x\\ &= e & & \text{substituting }x*x\inv=e \end{align*}
      From the previous part we see that \(e\) is the identity element of \(G^{op}\text{,}\) making \(x\inv\) the inverse of \(x\) in \(G^{op}\text{.}\) Thus every element in \(G^{op}\) has an inverse in \(G^{op}\text{,}\) satisfying the inverse element property.
  2. Integer addition is associative, which means that for any three integers \(a, b\text{,}\) and \(c\text{,}\) the expression \((a + b) + c\) is equal to \(a + (b + c)\text{.}\) Thus, associativity is satisfied for \(\E\text{.}\)
    The identity element for addition is \(0\text{.}\) For any even integer \(a\text{,}\) adding \(0\) to it does not change its parity (evenness). Hence, the identity element property is satisfied for \(\E\text{.}\)
    For every even integer \(a\text{,}\) there exists an inverse element denoted as \(-a\) such that \(a + (-a) = (-a) + a = 0\text{.}\) The negative of an even integer is also an even integer. Adding an even integer to its negative results in \(0\text{,}\) which is the identity element. Therefore, the inverse element property is satisfied for \(\E\text{.}\)
    The identity element for addition is \(0\text{.}\) However, \(0\) is not an odd integer, so the identity element property is not satisfied for \(\O\text{.}\)

Proof.

  1. The element \((x\inv)\inv\) is defined to be the inverse of the element \(x\inv\text{.}\) I claim that \(x\) is also the inverse of \(x\inv\text{.}\)
    As \(x\inv\) is the inverse of \(x\text{,}\) by the inverse property we have \(xx\inv=e\) and \(x\inv x=e\text{,}\) making \(x\) the inverse of \(x\inv\text{.}\) Thus both \(x\) and \((x\inv)\inv\) are inverses of the element \(x\inv\text{.}\) As inverses are unique, we see \((x^{-1})^{-1} = x\text{.}\)
  2. The element \((xy)\inv\) is defined to be the inverse of the element \(xy\text{.}\) I claim that \(y\inv x\inv\) is also the inverse of \(xy\text{.}\) Observe
    \begin{align*} (xy)(y\inv x\inv) &= x(yy\inv)x\inv && \knowl{./knowl/xref/associative-property.html}{\text{Associative Property}} \\ &= x(e)x\inv && y\inv \text{ is the inverse of } y \\ &= (xe)x\inv && \knowl{./knowl/xref/associative-property.html}{\text{Associative Property}} \\ &= xx\inv && \knowl{./knowl/xref/identity-property.html}{\text{\(e\) is the identity element}}\\ &= e && x\inv \text{ is the inverse of } x. \end{align*}
    and
    \begin{align*} (y\inv x\inv)(xy) &= y\inv(x\inv x)y && \knowl{./knowl/xref/associative-property.html}{\text{Associative Property}} \\ &= y\inv(e)y && x\inv \text{ is the inverse of } x \\ &= (y\inv e)y && \knowl{./knowl/xref/associative-property.html}{\text{Associative Property}} \\ &= y\inv y && \knowl{./knowl/xref/identity-property.html}{\text{\(e\) is the identity element}}\\ &= e && y\inv \text{ is the inverse of } y. \end{align*}
    Thus both \((xy)\inv\) and \(y\inv x\inv\) are inverses of \((xy)\text{,}\) making them equal as inverses are unique.
The second portion of Theorem 1.21 can be generalized.

Exercise 1.22. Big ’Ole Inverses.

If \(G\) is a group and \(x_1,\ldots,x_n \in G\text{,}\) then \((x_1 \dots x_n)^{-1} = x_n^{-1} \dots x_1^{-1}\text{.}\)

Exercise 1.23. Generalized Associative Law.

If an element of a group is contructed from a sequence of elements \(x_1,\dots,x_n\) in this order by repeatedly inserting parenthesis and applying the operation, the element must equal
\begin{equation*} (\cdots((x_1x_2)x_3)\cdots)x_n \end{equation*}
and so is independent of the mode of bracketing.
As a result of Exercise 1.23 any expression formed from the elements \(x_1\dots,x_n\) in that order can be written without parenthesis, which will save us oodles of time down the line.
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This is usually a result that is simply taken for granted and left unstated; we leave it here for the sake of completeness and ~rigor~. And yes, oodles is the best word to go here and no, I will not be changing it.
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