    # Groups II. Properties.

We shall defend our island, whatever the cost may be. We shall fight on the beaches, we shall fight on the landing-grounds, we shall fight in the fields and in the streets, we shall fight in the hills. We shall never surrender!, Winston Churchill.

# Order of Groups. Order of an element.

When working with generic groups, multiplication notation is very often used: gh (instead of g+h or g○h), gn=g·g···g (n times), and g-n = g-1g-1···g-1 (n times), and 1 for the identity (e). Definition. A group is finite if it contains a finite number of elements. Otherwise, it is said to be infinite. The number of elements in a group G is called the order of the group, and is denoted by |G|, e.g., ℤ10 and U10 = {1, 3, 7, 9} -positive integers less than 10 that are coprime to 10- are finite groups of orders 10 and 4 respectively. ℤn and Sn are finite groups of order n and n! respectively. Futhermore, (ℤ, +), (ℚ, +), (ℝ, +), and (ℂ, +) are infinite groups under addition.

When we are talking about the order of a group, we are really referring to the cardinality of the underlying set.

The order of an element a ∈ G is the least positive power of the element which gives us the identity, i.e., the least positive integer n such that an = 1, e.g., ord(3) = |3| = 2 in ℤ6 (3, 6) and ord(4) = |4| = 5 in ℤ10 (4, 8, 2, 6, 0). If no positive power of x is the identity, x is said to be of infinite order.

Examples:

• In (ℤ, +), (ℚ, +), (ℝ, +), and (ℂ, +), every non-zero element has infinite order.
• |0| = 1, |1| = 6, |2| = 3, |3| = 2, |4| = 3, |5| = 6 in ℤ6.
• |1| = |5| = |7| = |11| = 12, |2| = |10| = 6, |3| = |9| = 4, |4| = |8| = 3, and |6| = 2 in ℤ12.
• (ℝx, ·), |1| = 1, |-1| = 2, |a| = ∞, ∀a ≠ ±1.

An element of a group has order 1 if and only if it is the identity.

• In (ℂx, ·), there are infinite complex numbers that yield 1 when raised to some positive integer power n, zn = 1. They are called the roots of unity. In particular, i = $\sqrt{-1}$, |i| = 4.
• f ∈ S6, $f = (\begin{smallmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 1 & 3 & 2 & 5 & 4\end{smallmatrix}),f^2=(\begin{smallmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 4 & 6 & 3 & 1 & 5 & 2\end{smallmatrix}),f^3=(\begin{smallmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 4 & 3 & 6 & 5 & 1\end{smallmatrix}), f^4=(\begin{smallmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 2 & 3 & 4 & 5 & 2\end{smallmatrix})$, so |f| = 4.
• GL(2, ℝ) = $\bigl\{ {{[\bigl(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr)]: a, b, c, d ∈ ℝ, ad - bc ≠ 0}} \bigr\}$ is the group of invertible 2 x 2 matrices with entries in ℝ and non-zero determinant.

$A = (\begin{smallmatrix}-1 & -1\\ 1 & 0\end{smallmatrix}), A^2=(\begin{smallmatrix}-1 & -1\\ 1 & 0\end{smallmatrix})(\begin{smallmatrix}-1 & -1\\ 1 & 0\end{smallmatrix})=(\begin{smallmatrix}0 & 1\\ -1 & -1\end{smallmatrix}), A^3=A·A^2=(\begin{smallmatrix}-1 & -1\\ 1 & 0\end{smallmatrix})(\begin{smallmatrix}0 & 1\\ -1 & -1\end{smallmatrix})=(\begin{smallmatrix}1 & 0\\ 0 & 1\end{smallmatrix})$ = I2 ⇒ |A| = 3.

# Solving Equations in Groups

• Let’s solve 2x +1 = 5 in ℤ7 ↭ 2x + 1 ≡ 5 (mod 7) ⇒ [Existence of inverses] (2x + 1) + (-1) ≡ 5 + (-1) (mod 7) ⇒ [Associative] 2x + (1 + (-1)) ≡ 4 (mod 7) ⇒ [Existence of addition identity] 2x ≡ 4 (mod 7)

Notice that 2 ∈ U(7) because gcd(2, 7) = 1, so it has an inverse. 4·2 + (-1)·7 = 1 ⇒ 4·2 ≡ 1 (mod 7) ⇒ 4 = 2-1.

(U(7), ·) group, 2x ≡ 4 (mod 7) ⇒ [We are going to multiply both terms by the multiplicative inverse of 2, 2-1 = 4] 4·(2x) ≡ 16 (mod 7) ⇒ [Associative] (4·2)x ≡ 16 (mod 7) ⇒ 8x ≡ 16 (mod 7) ⇒ x ≡ 2 (mod 7).

• Suppose a, b, c ∈ G and G is an Abelian group. Solve axb = c.

axb = c ⇒ [G inverses] (axb)b-1 = cb-1 ⇒ [Associative property] (ax)(bb-1) = cb-1 ⇒ ax = cb-1 ⇒ [G inverses] a-1(ax) = a-1(cb-1) ⇒ [G associative] (a-1a)x = a-1(cb-1) ⇒ x = a-1(cb-1)

If G is Abelian ⇒ x = a-1(cb-1) = [G commutative and associative] (a-1b-1)c = [The shoe and sock principle. Let (G, ·) be a group. For any g, h ∈ G, (gh)-1 = h-1g-1] (ba)-1c, therefore x = (ba)-1c.

# Properties of Groups

Uniqueness of the identity. Let (G, ◦) be a group, then there is exactly one neutral element or identity e ∈ G such that eg = ge = g ∀g∈ G.

Proof. Let e, e’∈ G be neutral elements. Thus, ∀a ∈ G, we have ae = ea = a (i), ae’ = e’a = a (ii)
In particular, e’∈ G, e’e = ee’ = e’ (i).
Obviously e ∈ G, ee’ = e’e = e (ii). Thus, e = ee’ = e'.

Uniqueness of Inverses. Let (G, ◦) be a group, then there is a unique element b in G such that ab = ba = e.

Proof. Suppose a ∈ G, and let a-1, a’-1 be inverses of a, a-1 ≠ a’-1, aa-1=a-1a=e and aa’-1=a’-1a=e

aa-1 = a-1a ⇒ [Multiplying by a’-1 in both sides of the equation] a’-1(aa-1) = a’-1(a-1a) ⇒ [Associativity] (a’-1a)a-1 = a’-1(a-1a) ⇒ ea-1 = a’-1e ⇒ [Identity] a-1 = a’-1.

Left inverse for all is right inverse. Let (G, ◦) be a semigroup (it is closed under the group operation and associative) with an identity e such that, ∀a ∈ G, ∃b ∈ G: ba = e, that is, every element of S has a left inverse. Then, ab = e, that is, b is also a’s right inverse.

Proof. By assumption ba = e ⇒ [Multiplying by b in both sides of the equation] (ba)b = eb ⇒ [Associativity & Identity] b(ab) = b ⇒ [b has a left inverse, b-1] b-1b(ab) = b-1b = e ⇒ ab = e.

Proposition. The inverse of the inverse of an element is the element itself, that is (a-1)-1 = a.

Proof. This is just reinterpreting the equation aa-1 = a-1a = e and considering that inverses are unique in groups.

Cancellation property. Let (G, ·) be a group and let a, b, and c ∈ G. Then, the following statements hold true:
Left cancellation law. If ab = ac ⇒ b = c.
Right cancellation law. If ba = ca ⇒ b = c.

Proof. ab = ac ⇒ [∃a-1: aa-1 = a-1a = e. Multiplying both sides of the equation by a-1] a-1(ab) = a-1 (ac) ⇒ [Associativity] (a-1a)b = (a-1a)c ⇒ b = c.

The proof of the right cancellation law is completely analogous.

The shoe and sock principle. Let (G, ·) be a group. For any two arbitrary elements of the group a, b ∈ G, then (ab)-1 = b-1a-1.

When you get dressed, you first put your socks on, then your shoes. When you get naked, then the order is reverse. First, you take your shoes off, and then you take your socks off.

The result can be easily generalized, ∀a1, a2, ···, an, with inverses a1-1, a2-1, ···, an-1, then (a1a2···an)-1 = an-1···a2-1a1-1.

Proof.

Since inverses are unique in groups, it suffices to show that b-1a-1 acts like an inverse to (ab)-1.

ab(b-1a-1) = [Associativity] a(bb-1)a-1 = aa-1 = e.

(b-1a-1)ab = [Associativity] b-1(a-1a)b = b-1b = e.

Proposition. Let (G, ·) be a group and let a and b be any two arbitrary elements in G. Then, the equations ax = b and xa = b have unique solutions in G.

Proof: First, let’s show the existence of at least one solution. a-1b is a solution of ax=b. We leave the proof of xa=b as an exercise (it is quite similar).

a(a-1b) = [By associativity law] (aa-1)b = [By definition of inverses, a-1] eb = b.

To show uniqueness, let’s assume that we have two solutions y1 and y2.

ay1 = b = ay2 ⇒ ay1 = ay2 ⇒ [Multiplying by a-1 on both sides of the equation] a-1(ay1) = a-1(ay2) ⇒ [By Associativity law] (a-1a)y1 = (a-1a)y2 ⇒ [By definition of inverses, a-1] ey1 = ey2 ⇒ y1 = y2

Notation. Let G be a group and “a” an element of G, then we define a0 = e. For n ∈ ℕ, an = a·a···a, n times, and a-n = a-1·a-1···a-1, n times.

The Laws of exponents for groups. Let (G,⋅) be a group and let a ∈ G. If m and n are integers, then am⋅an=am+n, (am)n=amn, and (an)-1=(a-1)n.

• am⋅an = a·a··ₘ··a · a·a··ₙ··a = a·a··ₘ+ₙ··a = am+n.

• (am)n= a·a··ₘ··a · a·a··ₘ··a ··· a·a··ₘ··a, n copies of a·a··m··a = a·a··ₘₙ··a = amn.

• (an)-1 = (a-1)n. Let’s prove this using induction.

For the case base, n = 1, we have a·a-1 = a-1·a = e ⇒ (a-1)1 = a-1 = (a1)-1.

Suppose a-n = (an)-1 ⇒ an+1a-(n+1) = [By definition, an+1 = a·a··n+1·a =an·a, a-n = a-1·a-1··n··a-1, then a-(n+1) = a-1·a-1··n+1··a-1 = (a-1·a-1··n··a-1)·a-1 = a-n·a-1] aana-na-1 = [By Associativity] a(ana-n)a-1 = [By Induction hypothesis, a-n = (an)-1] aa-1 = e. Therefore, an+1a-(n+1) = e ⇒ (an+1)-1 = a-(n+1) = (a-1)n+1

Notation. When a group is Abelian, addition notation is often used: a+b as a·b or ab, 0 as e or 1, -a as the inverse of a (a-1), na instead of an = a + a +… + a (n times), and so on.

# Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.
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