1.1  Groups

Definition 1.1.1.

  1. A binary operation \(\whitestar\) on a set \(A\) is a function \(\whitestar\colon A\times A\to A\). We shall write \(x\whitestar y\) for \(\whitestar(x,y)\).

  2. A binary operation \(\whitestar\) on \(A\) is said to be associative if

    \[x\whitestar (y\whitestar z) = (x\whitestar y)\whitestar z\]

    for all \(x,y,z\in A\).

  3. Let \(\whitestar\) be a binary operation on \(A\). The elements \(x\) and \(y\) of \(A\) commute if

    \[x\whitestar y = y\whitestar x.\]

    The binary operation \(\whitestar\) is said to be commutative or abelian1 if \(x\) and \(y\) commute for all \(x,y\in A\).

Definition 1.1.2. Suppose that \(\whitestar\) is a binary operation on a set \(A\), and \(B\) is a subset of \(A\). If the restriction of \(\whitestar\) to \(B\) is a binary operation on \(B\), that is, \(x\whitestar y\in B\) for all \(x,y\in B\), then \(B\) is said to be closed under \(\whitestar\).

Note that if \(\whitestar\) is associative (respectively, commutative) on \(A\), and if \(B\) is closed under \(\whitestar\), then \(\whitestar\) is also associative (respectively, commutative) on \(B\).

Definition 1.1.3. A group is an ordered pair \((G,\whitestar)\) where \(G\) is a set and \(\whitestar\) is a binary operation on \(G\) satisfying the following axioms:

  1. the operation \(\whitestar\) is associative;

  2. there exists \(e\in G\), called an identity of \(G\), such that

    \[x\whitestar e = e \whitestar x = x\]

    for all \(x\in G\);

  3. for each \(x\in G\) there exists \(x^{-1}\in G\), called an inverse of \(x\), such that

    \[x\whitestar x^{-1} = x^{-1} \whitestar x = e.\]

Less formally, we shall often say that \(G\) is a group under \(\whitestar\). We also say the group \(G\) is commutative if \(\whitestar\) is commutative, and say \(G\) is finite if the set \(G\) is finite.

A group \(G\) under \(\whitestar\) may be viewed as a closed system of composable, reversible operations. Its binary operation \(\whitestar\) specifies how two elements are composed, while associativity ensures that successive compositions are unambiguous. The identity axiom guarantees the existence of an element that is neutral under composition on either side: composing it with any element leaves that element unchanged. The inverse axiom, in turn, guarantees that every element has a reverse element whose composition with it yields the identity.

Example 1.1.4.

  1. The sets \(\mathbb Z, \mathbb Q, \mathbb R\), and \(\mathbb C\) are groups under \(+\) with \(e=0\) and \(x^{-1} = -x\).
  2. The sets \(\mathbb Q\setminus \{0\}, \mathbb Q^+, \mathbb R\setminus \{0\}, \mathbb R^+\), and \(\mathbb C\setminus \{0\}\) are groups under \(\times\) with \(e=1\) and \(x^{-1} = 1/x\).

Definition 1.1.5. If \((G,\whitestar)\) and \((H,\diamond)\) are groups, we may form a new group \(G\times H\), called their direct product, whose elements are those in the Cartesian product \(G\times H\)

\[G \times H = \{(x,y) : x\in G, y\in H\}\]

and whose operation \(\circ\) is defined componentwise:

\[(x_1,y_1) \circ (x_2,y_2) = (x_1\whitestar x_2, y_1\diamond y_2).\]

We prove the uniqueness of the identity and the inverse of an element.

Proposition 1.1.6. If \(G\) is a group under \(\whitestar\), then

  1. the identity of \(G\) is unique;
  2. for each \(x\in G\), the inverse of \(x\) is unique;
  3. for all \(x\in G\), we have \((x^{-1})^{-1} = x\);
  4. for all \(x,y\in G\), we have \((x\whitestar y)^{-1} = y^{-1}\whitestar x^{-1}\);
  5. for any \(x_1,\ldots,x_n\in G\), the value of \(x_1\whitestar \cdots \whitestar x_n\) is independent of how the expression is bracketed.

Proof. (1) Let \(e,e'\in G\) be identities. Since \(e\) is an identity, \(e' = e\whitestar e'\). Since \(e'\) is an identity, we also have \(e = e\whitestar e'\). Therefore \(e=e'\). (2) Let \(y,z\in G\) be inverses of \(x\). Since \(e = x\whitestar z\), we have

\[y = y\whitestar e = y\whitestar (x\whitestar z).\]

Associativity of \(\whitestar\) gives

\[y\whitestar (x\whitestar z) = (y\whitestar x) \whitestar z.\]

Since \(y\whitestar x= e\),

\[(y\whitestar x) \whitestar z = e\whitestar z = z.\]

Therefore \(y=z\). (3) Let \(y=x^{-1}\). Then \(x\whitestar y = e\). Thus \(x=y^{-1}\). (4) Let \(z = (x\whitestar y)^{-1}\). Then \((x\whitestar y)\whitestar z = e\). Hence \(x^{-1} = y\whitestar z\). Therefore

\[y^{-1}\whitestar x^{-1} = y^{-1}\whitestar (y\whitestar z) = (y^{-1}\whitestar y)\whitestar z = e\whitestar z = z.\]

(5) \(\square\)

Henceforth, except when necessary, the product \(x\whitestar y\) will be written as \(xy\). Also, for any \(n\in \mathbb Z^+\), we write \(x^{n}\) and \(x^{-n}\) in place of

\[\underbrace{x x \cdots x}_{n\text{ times}}\quad \text{and} \quad \underbrace{x^{-1} x^{-1} \cdots x^{-1}}_{n\text{ times}}.\]

Of course, when we are dealing with specific groups, we shall use the natural notations.

Proposition 1.1.7. Let \(G\) be a group and let \(a,b\in G\). The equation \(ax=b\) and \(ya=b\) have unique solutions for \(x,y\in G\). In particular, the left and right cancellation law hold in \(G\), that is, if \(au=av\) then \(u=v\), and if \(ub =vb\) then \(u=v\).

Proof. If \(ax = b\) then \(x = a^{-1}b\). Since \(a^{-1}\) is unique, the solution \(x\) is unique. Similarly, if \(ya=b\) then \(y=ba^{-1}\). If \(au=av\) then \(u = a^{-1}av = ev = v\). Similarly, the right cancelation law holds.\(\square\)

The successive powers \(e,x,x^2,\ldots\) record what happens when an element \(x\) is repeatedly composed with itself. The first time this sequence returns to the identity leads to the notion of the order of an element.

Definition 1.1.8. An element \(x\) of a group \(G\) has finite order if

\[x^n = e\]

for some positive integer \(n\). In this case the order \(|x|\) is the smallest positive integer \(n\) such that \(x^n = e\). If there is no such \(n\), the order of \(x\) is defined to be infinity, and we write \(|x| = \infty\).

Definition 1.1.9. Let \(G = \{ x_1,x_2,\ldots,x_n\}\) be a finite group with \(x_1 = e\). The group table of \(G\) is the \(n\times n\) matrix whose \(i,j\) entry is the element \(x_ix_j\).

Exercises

1. Let \(G = \{x\in \mathbb R : 0\le x<1\}\) and let \(x\whitestar y = x + y - \lfloor x+y\rfloor\) for \(x,y\in G\). Prove that \(\whitestar\) is a binary operation on \(G\) and that \(G\) is an commutative group under \(\whitestar\). (\(G\) is called the real numbers modulo \(1\).)

Proof. Since \(a-1< \lfloor a \rfloor \le a\) for all \(a\in \mathbb R\), we have

\[0\le x\whitestar y = x + y - \lfloor x+y\rfloor < 1.\]

So \(\whitestar\) is a binary operation on \(G\). The operation \(\whitestar\) may be understood as a procedural operation adding two elements and then subtracting the integer part of the sum. Let \(x,y,z\in G\) and let \(\lfloor x+y\rfloor=n\). Then,

\[\begin{align} (x\whitestar y)\whitestar z &= (x+y-n)\whitestar z\\ &= ((x+y-n)+z) - \lfloor (x+y-n)+z\rfloor \\ &= (x+y+z) - \lfloor x+y+z\rfloor. \end{align}\]

Similarly,

\[x\whitestar (y\whitestar z) = (x+y+z) - \lfloor x+y+z\rfloor.\]

Therefore \(\whitestar\) is associative. Since

\[\begin{align} x\whitestar 0 &= (x+0) -\lfloor x+0\rfloor\\ &= x -\lfloor x\rfloor \\ &= x \end{align}\]

and \(0\whitestar x = x\) as well, \(0\) is the identity. If \(x=0\), its inverse is \(0\). If \(0<x<1\), put \(y=1-x\in G\). Then

\[x\whitestar y = 1-\lfloor 1\rfloor=0=y\whitestar x.\]

Therefore \(G\) is a group under \(\whitestar\). Furthermore, we have

\[\begin{align} x\whitestar y &= (x+y) -\lfloor x+y\rfloor\\ &= (y+x) -\lfloor y+x\rfloor\\ &= y\whitestar x. \end{align}\]

Consequently, the group \(G\) is commutative.\(\square\)

2. Let \(G = \{z\in \mathbb C : z^n = 1\text{ for some } n\in \mathbb Z^+\}\).

  1. Prove that \(G\) is a group under multiplication. (\(G\) is called the group of roots of unity in \(\mathbb C\).)
  2. Prove that \(G\) is not a group under addition.

Proof. (1) Let \(w,x,y,z\in G\). If \(z^n=1\) and \(w^m=1\), then \((zw)^{mn} = (z^n)^m(w^m)^n = 1\). Let \(x = (x_1,x_2), y=(y_1,y_2)\), and \(z=(z_1,z_2)\). Since we have

\[\begin{align} (xy)z &= (x_1y_1-x_2y_2,x_1y_2+x_2y_1)(z_1,z_2)\\ &= ((x_1y_1-x_2y_2)z_1-(x_1y_2 + x_2y_1)z_2, (x_1y_1-x_2y_2)z_2+ (x_1y_2 + x_2y_1)z_1)\\ &= (x_1y_1z_1-x_1y_2z_2-x_2y_1z_2-x_2y_2z_1, x_1y_1z_2 + x_1y_2z_1 +x_2y_1z_1 -x_2y_2z_2)\\ &= (x_1(y_1z_1 -y_2z_2) -x_2(y_1z_2 + y_2z_1),x_1(y_1z_2 + y_2z_1) + x_2(y_1z_1 -y_2z_2))\\ &= x(yz), \end{align}\]

multiplication is associative. Also, \(z1 = (z_1,z_2)(1,0) = (z_1,z_2) =z\) and \(1z = (1,0)(z_1,z_2) = (z_1,z_2) = z\). Hence \(1\) is the identity. For any \(z\in G\), we may let \(z^n = 1\). Then \(zz^{n-1} = 1\). Hence \(z^{n-1}\) is the inverse of \(z\). Therefore, \(G\) is a group under multiplication. (2) Since \(1^2=1\), \(1\) is an element of \(G\). However, \(1+1=2\notin G\), because there is no \(n\in \mathbb Z^+\) such that \(2^n=1\). Thus, \(G\) is not closed under addition.\(\square\)

3. Let \(x\) be an element of finite order in a group \(G\). Prove that if \(x^n = e\) for some positive integer \(n\), then \(|x|\) is a divisor of \(n\).

Proof. Let \(|x| = m\) and let \(n = mk + r\) where \(k,r\in \mathbb Z\) with \(k\ge 0\) and \(1\le r \le m\). Since \(x^m = e\), we have \(x^r = e\). By Definition 1.1.8, \(r\ge m\). Hence \(r = m\) and \(n = m(k+1)\). \(\square\)

4. Let \(G\) be a group and \(x\in G\). Prove that \(x\) and \(x^{-1}\) have the same order.

Proof. For every \(k\in\mathbb Z^+\), \((x^{-1})^k=(x^k)^{-1}\), so \((x^{-1})^k=e\) if and only if \(x^k=e\). Thus \(x\) and \(x^{-1}\) have the same order, whether finite or infinite.\(\square\)

5. Let \(G\) be a group and let \(x\in G\) be of order \(n\). Prove that if \(n\) is odd, then \(x = x^{2k}\) for some \(k\in \mathbb Z^+\).

Proof. Let \(n=2k-1\) for some \(k\in \mathbb Z^+\). Since \(x^n = e\), we have \(x^{n+1} = x\). Thus \(x^{2k} = x\).\(\square\)

6. Let \(G\) be a group and \(x,y\in G\). Prove that \(|x| = |y^{-1}xy|\), and deduce \(|ab|=|ba|\) for all \(a,b\in G\).

Proof. Let \(z=y^{-1}xy\). If \(z^n=y^{-1}x^ny\) for some \(n\in \mathbb Z^+\), then \(z^{n+1} = (y^{-1}x^ny)(y^{-1}xy) = y^{-1}x^{n+1}y\). By mathematical induction, we have \(z^n=y^{-1}x^ny\) for all \(n\in \mathbb Z^+\). Hence \(z^n=e\) if and only if \(x^n=e\) for every \(n\in\mathbb Z^+\). Therefore \(|x|=|z|\), whether finite or infinite. Taking \(xy=a\) and \(y^{-1}=b\), we obtain \(|ab|=|ba|\).\(\square\)

7. Let \(G\) be a group and \(x\in G\). If \(|x|=pq\) for some positive integers \(p\) and \(q\), prove that \(|x^p|= q\).

8. Let \(G\) be a group. Prove that if \(x^2=e\) for all \(x\in G\) then \(G\) is commutative.

  1. In the terms abelian group, “Abel” refers to the Norwegian mathematician Niels Henrik Abel (1802–1829). Despite dying at only 26, Abel made fundamental contributions to algebra and analysis. He is especially famous for proving that there is no general solution by radicals for polynomial equations of degree five, and for his work on elliptic functions and what are now called Abelian functions. His surname is pronounced /ˈeɪbəl/. ↩