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Subgroups.

Definition. Let G be a group. A subgroup is a subset of a group that itself is a group. Formally, H ⊂ G (H is a subset of G), ∀a, b ∈ H, a·b ∈ H, and (H, ·) is a group.

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

All other subgroups are said to be nontrivial or proper subgroups, H < G.

Examples: In ℤ9 under the operation +, the subset {0, 3, 6} forms a proper subgroup. 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).

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} are not subgroups because 1 + 1 = 2 ∉ {0, 1}, 3 + 3 = 2 ∉ {0, 3}, and 1 + 2 = 3 ∉ {0, 1, 2}.  

Subgroup Tests

Two-Step Subgroup Test. Let G be a group, a non empty subset H ⊂ G is a subgroup (H ≤ G) iff

Proof:

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

Conversely, suppose (i) and (ii). Associative 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 a has an inverse element.

Example. 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, w = cosΦ + isinΦ. zw = cos(θ + Φ) + i sin(θ + Φ), zw ∈ $\Tau$ so it is closed.

We are using the theorem z, w ∈ ℂ, z = r(cosθ + i sinθ), w = s(cosΦ + isinΦ), then zw = rscos(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| = $\sqrt{a^{2}+(-b)^{2}} = 1$

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

Proof. If H is a subgroup, ∀a, b ∈ H, b-1∈ H, then a·b-1 ∈ H.

Conversely, suppose a non empty subset H ⊂ G, and ∀a, b ∈ H, a·b-1 ∈ H, then 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 a neutral element and a has an inverse element for all a in H.

∀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.

Example. Let (ℤ, +), r ∈ ℤ, rℤ = {r·n | n∈ ℤ} is the subgroup of all integers divisible by r. If a and b are divisible by r, then a+b and -a are also divisible by r.

Theorem. Finite Subgroup Test. Let G be a group, a non empty finite subset H ⊂ G closed under the operation of G (∀a, b ∈ H, a·b ∈ H) is a subgroup of G.

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

If a = e ⇒ 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). However, H is finite ⇒ ai = aj and i > j ⇒ ai-j = e ⇒ [i-j > 0, ai-j∈ H] e ∈ H, a-1 = ai-j-1 ∈ H because i-j-1 ≥ 0.

Proposition. If H is a subgroup of G, and L is a subgroup of H, then L is a subgroup of G.

Proposition. If H and L are subgroups of G, then H∩L is a subgroup of G.

Proof: ∀a, b ∈ H∩L, a, b ∈ H and a, b ∈ L ⇒ a·b-1 ∈ H, 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, the set ⟨a⟩={an | n ∈ Z} is a subgroup of G. We call it the cyclic subgroup generated by a. Proof: g, h ∈ ⟨a⟩, g = am, h = an for some m, n ∈ ℤ, gh = am+n, g-1 = a-n ∈ ⟨a⟩. In ℤ10, ⟨2⟩ = {0, 2, 4, 6, 8}.

Center of a Group

The center of a group, G, is the subset of elements that commute with every element of G.Z(G) = {a ∈ G | ab = ba ∀ b in G}.

The center of a group, Z(G), is a subgroup of G, Z(G)≤ G.

Proof: [Two-Step Subgroup Test]

  1. ∀ a, b, x ∈ Z(G), ¿(ab)x = x(ba)? (ab)x = (ba)x [a ∈ Z(G)] = [Associativity] b(ax) = [a ∈ Z(G)] b(xa) = [Associativity] (bx)a = [b ∈ Z(G)] (xb)a = [Associativity] x(ba)
  2. ∀ a, x ∈ Z(G), ax = xa, ¿a-1x = xa-1? ∀ a, x ∈ Z(G), ax = xa. Let’s take a-1(ax) ∈ G, therefore, a((a-1)(ax)) = ((a-1)(ax))a

a((a-1)(ax)) = (aa-1)(ax) = e(ax) = ax
((a-1)(ax))a = (a-1a)(xa) = e(xa) = xa ∎

Examples.

x ∈ ℤ(Dn) iff xα = αx and xβ = βx.

x = αiβj ↔ αiβjα = ααiβj = αi+1βj ↔ βjα = αβj

j = 1, βα = αβ [βαβ = α-1 ⇒ βα = α-1β-1 = αβ] α-1β = αβ ⇒ α-1 = α ⇒ α2 = e ⊥. Therefore, x cannot be a reflection (all reflections are generated by αiβ, 0 ≤ i ≤ n-1), it necessary needs to be a rotation, x = αi.

x ∈ ℤ(Dn) ⇒ xβ = βx ⇒ αiβ = βαi ⇒ [αkβ = βαn-k] αiβ = αn-iβ = α-iβ ⇒ αi = α-i ⇒ α2i = e ⇒ i = 0 (x = α0 = e) or i = n/2 and n is even.

Therefore, ℤ(Dn) = $ \begin{cases} e, n~is~ odd\\\\ e~ and~ α^{\frac{n}{2}}, n~is~ even \end{cases}$

The Product HK of Two Subgroups H and K of a Group G

Definition. Let G be a group, and let H and K be two subgroups of G. The Product of H and K is HK = {hk: h ∈ H and k ∈ K}

There are cases where G is a group, H and K are two subgroups, and yet HK is not a subgroup of G. Example. Let G = S3, H = ⟨(1, 2)⟩, and K = ⟨(2, 3)⟩. Notice that H = {(1, 2), id} and K = {(2, 3), id}

HK = {id, (1, 3, 2), (1, 2), (2, 3)}. |HK| = 4 is not a divisor of 6 (=|G|), so HK is not a subgroup of G by Lagrange’s Theorem.

Proposition. Let G be a group and let H and K be two subgroups of G. Then, HK is a subgroup of G (HK ≤ G) if and only if HK = KH.

Proof. Suppose that HK is a subgroup. ∀x ∈ HK, ∃h ∈ H, ∃k ∈ K, x = hk. We claim that x ∈ KH.

x ∈ HK and HK is a subgroup ⇒ x-1 = (hk)-1 = k-1h-1 ∈ HK. However, x-1 = k-1h-1 ∈ KH because k-1 ∈ K (K ≤ G) and h-1 ∈ H (H ≤ G), therefore (x-1)-1 = x ∈ KH (KH ≤ G) ⇒ HK ⊆ KH.

∀x ∈ KH, ∃h ∈ H, k ∈ K, x = kh. We claim that x ∈ HK.

x = kh = (ek)(he) ∈ HK, because HK ≤ G, so there is closure and ek ∈ HK (e ∈ H), and he ∈ HK (e ∈ K). KH ⊆ HK ⇒ KH = HK.

Suppose that HK = KH. Is HK ≤ G? Let x, y ∈ HK, ∃h1, h2 ∈ H, k1, k2 ∈ K, x = h1k1 and y = h2k2.

xy = h1k1h2k2 = h1(k1h2)k2 = [HK = KH, ∃k’∈ K and h’ ∈ H] h1(h’k’)k2 ∈ HK because h1h’ ∈ H and k’k2 ∈ K, there is closure.

Associativity is inherited from G. e = e·e ∈ HK (identity). ∀x ∈ HK, ∃h ∈ H, ∃k ∈ K, x = hk. x-1 = k-1h-1 = [HK = KH, ∃k’∈ K and h’ ∈ H] h’k’ ∈ HK (inverses).

Proposition. Let G be a group and let H and K be two finite subgroups of G. Then $|HK|=\frac{|H||K|}{|H∩K|}$

Proof:

hk = h’k’ ⇒h’-1h = k’k-1 = t ∈ H∩K

k’k-1 = t ⇒ k’ = tk ⇒ k = t-1k’. h’-1h = t ⇒ h = h’t

Therefore, hk = h’k’ = (h’t)(t-1k’) for some t ∈ H∩K, so every element in HK can be written in exactly |H∩K| different ways as an element from H times an element from K, and so, the equality $|HK|=\frac{|H||K|}{|H∩K|}$ holds.

In particular, |H∩K| divides |H||K|.

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