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.

- (ℤ, +) ≤ (ℚ, +). (ℚ, +) ≤ (ℝ, +). The set of even integers is also a subgroup of (ℤ, +).
- In ℤ
_{9}under the addition operation +, the subset {0, 3, 6} forms a proper subgroup. Besides {0, 2, 4} ≤ ℤ_{6}and {0, 3} ≤ ℤ_{6}. However, {0, 1, 2, 3, 4} ≴ ℤ_{6}😞 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, r^{2}, ···, r^{n-1}, namely e^{2π/n}, e^{2(2π/n)}, ···, e^{(n-1)2π/n}. If {a_{1}, a_{2}, ···, a_{n}} denotes the vertices of a n-gon, the cyclic permutation α = (a_{1}, a_{2}, ···, a_{n}) ∈ S_{n} is the rotation r. Chose any vertex, e.g. a_{1}, and consider it the fixed point of a reflection of the plane, i.e., the permutation β = (a_{2}, a_{n})(a_{3}, a_{n-1})··· ∈ S_{n}. Since D_{n} is generated by α and β ∈ S_{n} ⇒ D_{n} ≤ S_{n}

- Q
_{8}= {$e, \bar e, \bar j, j, \bar k, k, \bar i, i$} = {$(\begin{smallmatrix}1 & 0\\ 0 & 1\end{smallmatrix}),(\begin{smallmatrix}-1 & 0\\ 0 & -1\end{smallmatrix}),(\begin{smallmatrix}0 & 1\\ -1 & 0\end{smallmatrix}), (\begin{smallmatrix}0 & -1\\ 1 & 0\end{smallmatrix}), (\begin{smallmatrix}0 & i\\ i & 0\end{smallmatrix}), (\begin{smallmatrix}0 & -i\\ -i & 0\end{smallmatrix}), (\begin{smallmatrix}-i & 0\\ 0 & i\end{smallmatrix}), (\begin{smallmatrix}i & 0\\ 0 & -i\end{smallmatrix})$} ≤ GL_{2}(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, Q_{8}≤ GL_{2}(ℂ).

Using matrix multiplication, we have Q_{8} = {±1, ±i, ±j, ±k} and i^{2} = j^{2} = k^{2} =−1, ij = −ji = k, jk = −kj = i, ki = −ik = j. Moreover, 1 is the identity of Q_{8}, 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**= {A ∈ GL_{n}(ℝ)_{n}(ℝ) | det(A)=1 }**≤ GL**._{n}(ℝ)

- Closure: ∀A, B ∈ SL
_{n}(ℝ): det(AB) = det(A)·det(B) = 1·1 = 1 ⇒ AB ∈ SL_{n}(ℝ) - det(I
_{n}) = 1, so I_{n}∈ SL_{n}(ℝ) - ∀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 ℤ**(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}._{4}is {0, 2}

**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^{-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 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^{iθ}e^{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, r^{2}, ···, r^{n-1}, namely e^{2π/n}, e^{2(2π/n)}, ···, e^{(n-1)2π/n}. Let H be {1, r, r^{2}, ···, 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, a^{2}, a^{3},… belong to H because H is closed under the operation of G (∀a, b ∈ H, a·b ∈ H, so a·a = a^{2} ∈ H and so on). However, since H is finite, this process will necessary produce repetitions ⇒∃i, j: a^{i} = 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⟩={a ^{n} | n ∈ Z} is the cyclic subgroup generated by a**.

Proof: a^{0} = e ∈ ⟨a⟩. g, h ∈ ⟨a⟩, g = a^{m}, h = a^{n} for some m, n ∈ ℤ, gh = a^{m+n} ∈ ⟨a⟩, m+n ∈ ℤ; g^{-1} = a^{-m} ∈ ⟨a⟩ (a^{m}a^{-m} = a^{0} = e, -m ∈ ℤ)∎

Examples:

- ⟨2⟩ = {0, 2, 4, 6, 8} ≤ ℤ
_{10}. - In ℤ
_{12}, ⟨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, r
^{2}, ···, 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.

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.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts Mathematics, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.
- Michael Penn (Abstract Algebra), and MathMajor.
- Contemporary Abstract Algebra, Joseph, A. Gallian.
- Andrew Misseldine: College Algebra and Abstract Algebra.