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 ℤ_{9} 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 ℤ_{4} 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}.

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

- (i) H is
**closed**under the 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). Associative is inherited from the group product of G.

∀a ∈ H, a^{-1} ∈ H (ii) ⇒ a(a^{-1}) = (a^{-1}a) = 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, a^{2}, a^{3},… belong to H because H is closed under the operation of G (∀a, b ∈ H, a·b ∈ H). However, H is finite ⇒ a^{i} = a^{j} and i > j ⇒ a^{i-j} = e ⇒ [i-j > 0, a^{i-j}∈ H] e ∈ H, a^{-1} = a^{i-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⟩={a^{n} | n ∈ Z} is a subgroup of G. We call it the cyclic subgroup generated by a. Proof: g, h ∈ ⟨a⟩, g = a^{m}, h = a^{n} for some m, n ∈ ℤ, gh = a^{m+n}, g^{-1} = a^{-n} ∈ ⟨a⟩. In ℤ_{10}, ⟨2⟩ = {0, 2, 4, 6, 8}.

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]

- ∀ 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)
- ∀ a, x ∈ Z(G), ax = xa, ¿a
^{-1}x = 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^{-1}a)(xa) = e(xa) = xa ∎

Examples.

- The center of an Abelian group, G, is the group itself.
- Let’s calculate the center of the dihedral group D
_{n}= {α, β: α^{n}= β^{2}= e, βαβ = α^{-1}}, n ≥ 3.

x ∈ ℤ(D_{n}) iff xα = αx and xβ = βx.

x = α^{i}β^{j} ↔ α^{i}β^{j}α = αα^{i}β^{j} = α^{i+1}β^{j} ↔ β^{j}α = αβ^{j}

j = 1, βα = α^{}β [βαβ = α^{-1} ⇒ βα = α^{-1}β^{-1} = αβ] α^{-1}β = αβ ⇒ α^{-1} = α ⇒ α^{2} = e ⊥. Therefore, x cannot be a reflection (all reflections are generated by α^{i}β, 0 ≤ i ≤ n-1), it necessary needs to be a rotation, x = α^{i}.

x ∈ ℤ(D_{n}) ⇒ xβ = βx ⇒ α^{i}β = βα^{i} ⇒ [α^{k}β = βα^{n-k}] α^{i}β = α^{n-i}β = α^{-i}β ⇒ α^{i} = α^{-i} ⇒ α^{2i} = e ⇒ i = 0 (x = α^{0} = e) or i = n/2 and n is even.

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

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 = S_{3}, 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^{-1}h^{-1} ∈ HK. However, x^{-1} = k^{-1}h^{-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, ∃h_{1}, h_{2} ∈ H, k_{1}, k_{2} ∈ K, x = h_{1}k_{1} and y = h_{2}k_{2}.

xy = h_{1}k_{1}h_{2}k_{2} = h_{1}(k_{1}h_{2})k_{2} = [HK = KH, ∃k’∈ K and h’ ∈ H] h_{1}(h’k’)k_{2} ∈ HK because h_{1}h’ ∈ H and k’k_{2} ∈ 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^{-1}h^{-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:

- H ≤ G and K ≤ G ⇒ H∩K ≤ G ⇒ |H∩K| ≥ 1 because e ∈ H∩K, so we are never dividing by zero.
- ∀t ∈ H∩K, ∀h ∈ H, k ∈ K, hk = ht(t
^{-1}k) ∈ HK because ht ∈ H, t^{-1}k ∈ K, so each element in HK can be written by at least |H∩K| different ways as an element from H times and element from K.

hk = h’k’ ⇒h’^{-1}h = k’k^{-1} = t ∈ H∩K

k’k^{-1} = t ⇒ k’ = tk ⇒ k = t^{-1}k’. h’^{-1}h = t ⇒ h = h’t

Therefore, hk = h’k’ = (h’t)(t^{-1}k’) 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|.