Subgroups.

For example is not proof but it does show that something is possible.

Definition. Let G be a group. A subgroup is a subset of a group which is also a group of its own, that is, that follows all axioms that are required to form a group. Formally, H ⊂ G (H is a subset of G), ∀a, b ∈ H, a·b ∈ H, and (H, ·) is a group (closure, identity, and inverses).

We use the notation H ≤ G to mean that H is a subgroup of G. Every group G has at least two trivial subgroups: the group G itself and the subgroup {e} containing only the identity element. Notice that {e}, a set, is different than e, the identity element.

All other subgroups are said to be nontrivial or proper subgroups, and we express it as H < G.

Examples

• (ℤ, +) ≤ (ℚ, +). (ℚ, +) ≤ (ℝ, +). The set of even integers is also a subgroup of (ℤ, +).
• In ℤ₉ under the addition operation +, the subset {0, 3, 6} forms a proper subgroup. Besides {0, 2, 4} ≤ ℤ₆ and {0, 3} ≤ ℤ₆. However, {0, 1, 2, 3, 4} ≴ ℤ₆ 😞 because it is not closed under the group operation, e.g., 1 + 4 ∉ {0, 1, 2, 3, 4}).
• {1, -1} ≤ ℝ* and {1, -1, i, -i} ≤ (ℂ*, ·) are finite subgroups of order 2 and 4 respectively.
• The dihedral group is a subgroup of the symmetric group, D_n ≤ S_n.

Let r be the counterclockwise rotation by θ = $\frac{2π}{n}$, then the n rotations in D_n are e, r, r², ···, r^n-1, namely e^2π/n, e^2(2π/n), ···, e^(n-1)2π/n. If {a₁, a₂, ···, a_n} denotes the vertices of a n-gon, the cyclic permutation α = (a₁, a₂, ···, a_n) ∈ S_n is the rotation r. Chose any vertex, e.g. a₁, and consider it the fixed point of a reflection of the plane, i.e., the permutation β = (a₂, a_n)(a₃, a_n-1)··· ∈ S_n. Since D_n is generated by α and β ∈ S_n ⇒ D_n ≤ S_n

• Q₈ = {$e, \bar e, \bar j, j, \bar k, k, \bar i, i$} = {$(\begin{smallmatrix}1 & 0\\ 0 & 1\end{smallmatrix}),(\begin{smallmatrix}-1 & 0\\ 0 & -1\end{smallmatrix}),(\begin{smallmatrix}0 & 1\\ -1 & 0\end{smallmatrix}), (\begin{smallmatrix}0 & -1\\ 1 & 0\end{smallmatrix}), (\begin{smallmatrix}0 & i\\ i & 0\end{smallmatrix}), (\begin{smallmatrix}0 & -i\\ -i & 0\end{smallmatrix}), (\begin{smallmatrix}-i & 0\\ 0 & i\end{smallmatrix}), (\begin{smallmatrix}i & 0\\ 0 & -i\end{smallmatrix})$} ≤ GL₂(ℂ) = GL(2, ℂ), that is, the quaternion group is a finite, non-Abelian subgroup of the general linear group where e is the identity element, Q₈ ≤ GL₂(ℂ).

Using matrix multiplication, we have Q₈ = {±1, ±i, ±j, ±k} and i² = j² = k² =−1, ij = −ji = k, jk = −kj = i, ki = −ik = j. Moreover, 1 is the identity of Q₈, it is closed under matrix multiplication, and contains the inverse of every element, so it is a subgroup of GL(2, ℂ).

• GL_n(ℝ), the set of invertible n x n matrices with real entries, is a group under matrix multiplication. SL_n(ℝ), the set of n x n matrices with real entries whose determinant is equal to 1, is a proper subgroup of GL_n(ℝ). SL_n(ℝ) = {A ∈ GL_n(ℝ) | det(A)=1 } ≤ GL_n(ℝ).

1. Closure: ∀A, B ∈ SL_n(ℝ): det(AB) = det(A)·det(B) = 1·1 = 1 ⇒ AB ∈ SL_n(ℝ)
2. det(I_n) = 1, so I_n ∈ SL_n(ℝ)
3. ∀A ∈ SL_n(ℝ), ∃A^-1∈ SL_n(ℝ) such that A x A^-1 = I_n ⇒ det(A)·det(A^-1) = 1 ⇒ [det(A)=1] det(A^-1) = 1 ⇒ A^-1 ∈ SL_n(ℝ)∎

• Suppose H ≤ (ℤ, +) and H ≠ {0}. Let n ∈ H be the smallest positive number in H and m ∈ H be any other element in H. By the division algorithm, ∃q, r ∈ ℤ, 0 ≤ r < n: m = nq + r ⇒ r = m - nq = m -n -n ··· (q times) -n ∈ H ⇒ [r ∈ H, r < n, and by assumption, n ∈ H is the smallest positive number in H] r = 0 ⇒ m = nq ⇒ H = nℤ = {nk : k ∈ ℤ}.
• Let G = ℂ^* = {a + bi|a, b ∈ ℝ, not both zeroes}, then ℚ^* ≤ ℝ^* ≤ ℂ^*.
• The Klein four-group is a group with four elements, in which each element is self-inverse, that is, composing it with itself produces the identity (Figure 1.a.). It has three non-trivial proper subgroups {e, a}, {e, b}, and {e, c}. Notice that {e, a, b}, {e, a, c} or {e, b, c} are not subgroups because they are not closed: ab = c ∉ {e, a, b}, ac = b ∉ {e, a, c}, and bc = a ∉ {e, b, c}.
• The only nontrivial proper subgroup of ℤ₄ is {0, 2} (Figure 1.a.), e.g., {0, 1}, {0, 3}, {0, 1, 2}, and {0, 2, 3} are not subgroups because 1 + 1 = 2 ∉ {0, 1}, 3 + 3 = 2 ∉ {0, 3}, 1 + 2 = 3 ∉ {0, 1, 2}, and 2 + 3 = 1 ∉ {0, 2, 3}.  

Subgroup Tests

Two-Step Subgroup Test. Let G be a group, a non empty subset H ⊂ G is a subgroup (H ≤ G) iff (it is short for if and only if)

• (i) H is closed under the group operation in G, ∀a, b ∈ H, a·b ∈ H.
• (ii) Every element in H has an inverse in H, ∀a ∈ H, a^-1 ∈ H.

Proof:

If H is a subgroup, then (i) and (ii) obviously hold.

Conversely, suppose (i) and (ii). The associative o grouping property is inherited from the group product of G.

∀a ∈ H, a^-1 ∈ H (ii) ⇒ a(a^-1) = (a^-1a) = e ∈ H (i) ⇒ e is a neutral element of G, so it is a neutral element of H ⇒ H has a neutral element and every element in H (∀a ∈ H) has an inverse element.

Examples.

• The circle group, $\Tau$ = {z ∈ ℂ: |z|=1} is a proper subgroup of ℂ*. It is the multiplicate group of all complex numbers with absolute value 1, that is, the unit circle in the complex plane. It has infinite order.  

Proof. If z, w ∈ $\Tau$, z = cosθ + i sinθ = e^iθ, w = cosΦ + isinΦ = e^iΦ. zw = e^iθe^iΦ = e^{i(θ + Φ)} = cos(θ + Φ) + i sin(θ + Φ), zw ∈ $\Tau$ so it is closed.

This is basic stuff, z, w ∈ ℂ, z = r(cosθ + i sinθ), w = s(cosΦ + isinΦ), then zw = rs(cos(θ + Φ) + i sin(θ + Φ)).

If z ∈ $\Tau$, z = a + ib, r = |z| = $\sqrt{a^{2}+b^{2}} = 1$, z^-1 = $\frac{a-bi}{a^{2}+b^{2}} = a - bi,$ and r’ = |z^-1| = $\sqrt{a^{2}+(-b)^{2}} = 1$ ⇒ z^-1 ∈ $\Tau$.

• Let G = D_n be the dihedral group of order n, let r be the counterclockwise rotation by θ = $\frac{2π}{n}$, then the n rotations in D_n are e, r, r², ···, r^n-1, namely e^2π/n, e^2(2π/n), ···, e^(n-1)2π/n. Let H be {1, r, r², ···, r^n-1} the subset of all rotations in G. The product of two rotations is again a rotation, and the inverse of a rotation is also a rotation ⇒ H ≤ D_2n.

Theorem. One-Step Subgroup Test. Let G be a group, a non empty arbitrary subset of G, H ⊆ G, is a subgroup of G (H ≤ G) iff ∀a, b ∈ H, a·b^-1 ∈ H.

Proof. If H ≤ G, ∀a, b ∈ H ⇒ [H subgroup ⇒ inverses] b^-1∈ H ⇒[Closed operation] a·b^-1 ∈ H.

Conversely, suppose a non empty arbitrary subset of G, say H ⊆ G, and ∀a, b ∈ H, a·b^-1 ∈ H. In particular, take a = b ⇒ b·b^-1 = b^-1·b = e ∈ H ⇒ e is a neutral element of G, so it is a neutral element of H ⇒ H has G’neutral element e and any arbitrary element in H, say b ∈ H, has an inverse element, b^-1∈ G ⇒ [e, b ∈ H] e·b^-1 = b^-1 ∈ H.

Closed operation. ∀a, b ∈ H, let’s show that their product is also in H. We know that b^-1 ∈ H ⇒ (b^-1)^-1 ∈ H ⇒ a·(b^-1)^-1 = ab ∈ H ∎

Theorem. Finite Subgroup Test. Let G be a group,a non empty finite subset of G is a subgroup if and only if it is closed under the operation of G,i.e., H ⊆ G, |H| = n, such that ∀a, b ∈ H, a·b ∈ H ⇒ H ≤ G.

Proof. Using the Two-Set Subgroup Test, we only need to prove that a^-1 ∈ H, ∀a ∈ H.

If a = e ⇒ [By assumption, H is closed under the operation of G] ee = e ∈ H and a^-1 = e ∈ H.

If a ≠ e ⇒ a, a², a³,… belong to H because H is closed under the operation of G (∀a, b ∈ H, a·b ∈ H, so a·a = a² ∈ H and so on). However, since H is finite, this process will necessary produce repetitions ⇒∃i, j: aⁱ = a^j and i > j ⇒ a^i-j = e ∈ H ⇒ [∃i, j such that i-j > 0, a^i-j∈ H] Therefore, a^-1 = a^i-j-1 ∈ H because 0 ≤ i-j-1 < i-j and a · a^i-j-1 = a^i-j-1 · a = a^i-j = e∎

Proposition. Subgroup relation is transitive. If K is a subgroup of G, and H is a subgroup of K, then H is a subgroup of G, H ≤ K ≤ G ⇒ H ≤ G.

Proof.

H ≤ K ≤ G ⇒ H ⊆ K ⊆ G ⇒ H is a subset of G, H ⊆ G. The closure of H under G’s operation is not changed, whether you are looking at H as a subset of K or G under the same group operation.

Futhermore, since the identity element in G, say e_G, is unique ⇒ e_G = e_K = e_H. Similarity, ∀a ∈ H, a^-1 ∈ H (H ≤ K), and this inverse is also unique (Uniqueness of identity and inverses.) ∎

Proposition. Intersection of two subgroups of a group is again a subgroup. If H ≤ G, L ≤ G, then H∩L ≤ G.

Proof: ∀a, b ∈ H∩L, a, b ∈ H and a, b ∈ L ⇒ [H ≤ G, L ≤ G, and One-Step Test] a·b^-1 ∈ H and a·b^-1 ∈ L ⇒ a·b^-1 ∈ H∩L∎

Examples. Let G be a group and a be any element in G (a ∈ G). Then, a generating set of a group is a subset of the group such that every element of the group can be expressed as a combination (under the group operation) of finitely many elements of the subset and their inverses. In particular, the set ⟨a⟩={aⁿ | n ∈ Z} is the cyclic subgroup generated by a.

Proof: a⁰ = e ∈ ⟨a⟩. g, h ∈ ⟨a⟩, g = a^m, h = aⁿ for some m, n ∈ ℤ, gh = a^m+n ∈ ⟨a⟩, m+n ∈ ℤ; g^-1 = a^-m ∈ ⟨a⟩ (a^ma^-m = a⁰ = e, -m ∈ ℤ)∎

Examples:

• ⟨2⟩ = {0, 2, 4, 6, 8} ≤ ℤ₁₀.
• In ℤ₁₂, ⟨0⟩ = {0}, ⟨1⟩ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}, ⟨2⟩ = {0, 2, 4, 6, 10}, ⟨3⟩ = {0, 3, 6, 9}, ⟨4⟩ = {0, 4, 8}, ⟨6⟩ = {0, 6}.
• Let r be the counterclockwise rotation by θ = $\frac{2π}{n}$, H = {e, r, r², ···, r^n-1} = ⟨r⟩ ≤ D_n. Futhermore, ⟨r, s⟩ = D_n where s can be any reflection across a line of symmetry, that is, D_n is generated by both r and s.
• ⟨1⟩ = (ℤ, +), ⟨-1⟩ = ⟨1⟩ = ℤ.

Bibliography

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