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.
When working with generic groups, multiplication notation is very often used: gh (instead of g+h or g○h), g^{n}=g·g···g (n times), and g^{-n} = g^{-1}g^{-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., ℤ_{10} and U_{10} = {1, 3, 7, 9} -positive integers less than 10 that are coprime to 10- are finite groups of orders 10 and 4 respectively. ℤ_{n} and S_{n} 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 a^{n} = 1, e.g., ord(3) = |3| = 2 in ℤ_{6} (3, 6) and ord(4) = |4| = 5 in ℤ_{10} (4, 8, 2, 6, 0). If no positive power of x is the identity, x is said to be of infinite order.
Examples:
An element of a group has order 1 if and only if it is the identity.
$A = (\begin{smallmatrix}-1 & -1\\ 1 & 0\end{smallmatrix}), A^2=(\begin{smallmatrix}-1 & -1\\ 1 & 0\end{smallmatrix})(\begin{smallmatrix}-1 & -1\\ 1 & 0\end{smallmatrix})=(\begin{smallmatrix}0 & 1\\ -1 & -1\end{smallmatrix}), A^3=A·A^2=(\begin{smallmatrix}-1 & -1\\ 1 & 0\end{smallmatrix})(\begin{smallmatrix}0 & 1\\ -1 & -1\end{smallmatrix})=(\begin{smallmatrix}1 & 0\\ 0 & 1\end{smallmatrix})$ = I_{2} ⇒ |A| = 3.
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).
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^{-1}a)x = a^{-1}(cb^{-1}) ⇒ x = a^{-1}(cb^{-1})
If G is Abelian ⇒ x = a^{-1}(cb^{-1}) = [G commutative and associative] (a^{-1}b^{-1})c = [The shoe and sock principle. Let (G, ·) be a group. For any g, h ∈ G, (gh)^{-1} = h^{-1}g^{-1}] (ba)^{-1}c, therefore x = (ba)^{-1}c.
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^{-1}a=e and aa’^{-1}=a’^{-1}a=e
aa^{-1} = a^{-1}a ⇒ [Multiplying by a’^{-1} in both sides of the equation] a’^{-1}(aa^{-1}) = a’^{-1}(a^{-1}a) ⇒ [Associativity] (a’^{-1}a)a^{-1} = a’^{-1}(a^{-1}a) ⇒ ea^{-1} = a’^{-1}e ⇒ [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^{-1}b(ab) = b^{-1}b = 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^{-1}a = 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^{-1}a = e. Multiplying both sides of the equation by a^{-1}] a^{-1}(ab) = a^{-1} (ac) ⇒ [Associativity] (a^{-1}a)b = (a^{-1}a)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^{-1}a^{-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.
Proof.
Since inverses are unique in groups, it suffices to show that b^{-1}a^{-1} acts like an inverse to (ab)^{-1}.
ab(b^{-1}a^{-1}) = [Associativity] a(bb^{-1})a^{-1} = aa^{-1} = e.
(b^{-1}a^{-1})ab = [Associativity] b^{-1}(a^{-1}a)b = b^{-1}b = 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^{-1}b is a solution of ax=b. We leave the proof of xa=b as an exercise (it is quite similar).
a(a^{-1}b) = [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 y_{1} and y_{2}.
ay_{1} = b = ay_{2} ⇒ ay_{1} = ay_{2} ⇒ [Multiplying by a^{-1} on both sides of the equation] a^{-1}(ay_{1}) = a^{-1}(ay_{2}) ⇒ [By Associativity law] (a^{-1}a)y_{1} = (a^{-1}a)y_{2} ⇒ [By definition of inverses, a^{-1}] ey_{1} = ey_{2} ⇒ y_{1} = y_{2}
Notation. Let G be a group and “a” an element of G, then we define a^{0} = e. For n ∈ ℕ, a^{n} = 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 a^{m}⋅a^{n}=a^{m+n}, (a^{m})^{n}=a^{mn}, and (a^{n})^{-1}=(a^{-1})^{n}.
a^{m}⋅a^{n} = a·a··ₘ··a · a·a··ₙ··a = a·a··ₘ+ₙ··a = a^{m+n}.
(a^{m})^{n}= a·a··ₘ··a · a·a··ₘ··a ··· a·a··ₘ··a, n copies of a·a··_{m}··a = a·a··ₘₙ··a = a^{mn}.
(a^{n})^{-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} = (a^{1})^{-1}.
Suppose a^{-n} = (a^{n})^{-1} ⇒ a^{n+1}a^{-(n+1)} = [By definition, a^{n+1} = a·a··_{n+1}·a =a^{n}·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}] aa^{n}a^{-n}a^{-1} = [By Associativity] a(a^{n}a^{-n})a^{-1} = [By Induction hypothesis, a^{-n} = (a^{n})^{-1}] aa^{-1} = e. Therefore, a^{n+1}a^{-(n+1)} = e ⇒ (a^{n+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 a^{n} = a + a +… + a (n times), and so on.