Isomorphism theorems

Let H and K two subgroups of a group G. Their internal product is the subset of G definied by

The internal product need not be a subgroup, though at times it is.

For example if G = S₃, H = {id, (1,2)}, K = {id, (13)}, then HK = {id, (1,3), (1,2),(1,3,2)}. This subsets is not clearly a subgroup of S₃ (see Lagrange Theorem). Notice also as KH ≠ HK.

The following proposition holds true.

Proposition 7.13.1. Let G a group and N, H two subgroups of G such that N ⊲̲ G. Then NH = HN is a subgroup of G.

Proof. Being N normal in G, we have NH = HN. Let x = nh and y = n'h' two elements in NH: we prove that xy⁻¹ belongs to NH. It results

In the proof, it is not striclty necessary that N is normal it is sufficient that NH = HN to be a subgroup of G (see exercise 4). If G is abelian, the product HK of two of its arbitrary subgroups is a subgroup.

If H and K are two arbitrary subgroups of a group G, such that HK = KH, we just showed that HK is a subgroup. But then it coincides with the subgroup generated by H and K, that is

Proposition 7.13.2. Let φ a homomorphism between two group G and G'. Then

Corollary 7.13.3. Let G a group and N a normal subgroup in G; π: G ⟶ G/N the canonical projection. Then

1. If H is a subgroup of G, π(H) = HN/N is a subgroup of G/N;

2. If H' is a subgroup of G/N, π⁻¹(H') is a subgroup of G containing N.

Proof. It is sufficient to notice that the homomorphism π is surjective, hence π(G) = Im π = G/N.   □

Remark on the notation. Having defined the projection homomorphism π: G → G/N by g → gN; Then it is true that π(H) = π(HN). π(H) is a set of cosets of the form {hN} whilst π(HN) is the set of cosets {hnN} i.e. {hN}.

Proposition 7.13.4. Let G a group and let N ⊲̲ G. Let π the canonical projection on the quotient:

Then there exists a bijection Ψ between the subgroups H of G contaning N = ker π and the subgroups of the quotient G/N.

Proof. We defines Ψ as follows

We prove that ψ is invertible by showing that it is both injective and surjective:

1. π⁻¹(π(H)) = H   ∀H ≤ G,H ⊇ N.

2. π(π⁻¹(K)) = K   ∀K ≤ G/N.

It results always H ⊆ π⁻¹(π(H)). Let then x ∈ π⁻¹(π(H)); this means that π(x) ∈ π(H), that is π(x) = π(h) for some h ∈ H. But then x and h are related modulo N, that is x ∈ hN. But N ⊆ H, from which x ∈ H. Condition (ii) holds being π surjective.   □

Proposition 7.13.4. (First isomorphism theorem) Let G a group, N a normal subgroup of G and π the canonical projection. Let H an arbitrary subgroup of G then

1. π⁻¹(π(H)) = HN;

2. N ∩ H is normal in H.

3. H/(N ∩ H) ≃ HN/N

Proof. Let H ≤ G. Then

we have

The other inclusion holds as well: π⁻¹(π(H)) ⊆ HN, since if x ∈ π⁻¹(π(H)), then π(x) ∈ π(H), from which π(x) = π(h), that is x ∈ hN.

N ∩ H ⊲̲ H: let π_|H the restriction to H of the canonical projection π: G ⟶ G/N,

this is a homomorphsim with

Remember that N is the identity element of G/N so π_|H = {h ∈ H | hN = N } = H ∩ N.

These relations tell us that H ∩ N is normal in H, being the kernel of a homomorphism (theorem 1.12.1) and by the fundamental theorem of group homomorphism that (theorem 1.12.2)

Proposition 7.13.5. (Second isomorphism theorem) Let G a group and N ⊲̲ G. If H is a subgroup of G containing N, then

Moreover, it results

Proof. Let H ⊲̲ G, H ⊇ N. To show that H/N is normal in G/N we shall show that it is closed under conjugation: ∀ gN ∈ G/N and every hN ∈ H/N

Conversely, let now H/N ⊲̲ G/N. Consider the following compositions

where π₁ and π₂ are the canonical projection. The composition ρ = π₂ ∘ π₁ is an epimorphism with kernel H, hence H ⊲̲ G. By the fundamental theorem of group homomorphism

Example 7.13.5. Let G = Z₂₄, H = ⟨ [2]⟩, and N = ⟨[6]⟩, we have

then H/N = {([0] + N) + N, ([2] + N) + N ,([4] + N), ..., ([22] + N) + N} = {[0] + N, [2] + N, [4] + N, ..., [22] + N}, but the first three of those are the same as the next three (respectively) and then the same as the next three and then the same as the last three, because [0] + N = [6] + N = [12] + N = [18] + N, and [2] + N = [8] + N = [14] + N = [20] + N, etc.., so

Proposition 7.13.6. The alternating group A_n forms a normal subgroup of index 2 in S_n, that is (S_n: A_n) = 2.

Proof. The sign function sgn is a homomorphism from S_n onto (Z₂, +):

By definition of alternating group we have

Then ker(sgn) = A_n and therefore A_n is a normal subgroup of S_n from Theorem 7.13.2 and Proposition 7.12.10. Since Im(sgn) = Z₂ we have the sieze of the image is 2, that is |Im(sgn)| = 2 and hence |S_n/A_n| = 2. Therefore [S_n : A_n] = 2. Since |S_n| = n! then |A_n| = n!/2 follows from Lagrange's Thereom.   □

Exercises

1. Let G = GL₂(R) and let H = ⟨(1, 0; 1, 1)⟩ and K = ⟨(1, 1; 0, 1)⟩. Study H and K and verify that HK is not a subgroup of GL₂(R).

2. Let Q* the rational multiplicative group, N = {−1,1} and H the subgroup of Q* generated by 1/3. Study HN/N verifying the first fundamental isomorphism theorem.

Solutions

1. H = {(1, 0; 1, 1)^k = (1, 0; k, 1) | k ∈ Z}
   

   K = {(1, 1; 0, 1)^m = (1, 0; m, 1) | m ∈ Z}}

   This means that H and K are both cyclic infinite and thus by isomorphic to (Z,+). HK is a subgroup of G only if HK = KH (see proposition 7.13.1).

   A general element in HK is the product (1,0; k, 1)* (1, m; 0, 1) = (1, m; k, km + 1). A general element of KH is (1, m; 0, 1) * (1, 0; k, 1) = (1 + km, m; k, 1). The last element of a matrix in HK is always 1 in and one in KH has the 1st always 1 in. Thus HK ≠ KH.   ■

2. H = {(1/3)ⁱ| i ∈ Z}, HN = {hn | h ∈ H, n ∈ N} = {±(1/3)ⁱ| i ∈ Z}, H ∩ N = {1}. Then HN/N is all the cosets, xN, for x in HN, which is the same as all cosets xN for x in H, since hnN= hN for any h in H, any n in N. Thus x ± 1/3ⁱ. Then every class of HN/N is a power of (1/3)N and HN/N is a cyclic infinite gorup generated by the class (1/3)N. Since H ∩ N = {1}, then H/(H ∩ N) ≃ H, which is infinite cyclic. Then H/(H ∩ N) ≃ HN/N