Second Isomorphism Theorem

Statement

Let $G$ be a group, $N \trianglelefteq G$ a normal subgroup, and $H \leq G$ any subgroup. Then:

1. $HN = \{hn \mid h \in H,\, n \in N\}$ is a subgroup of $G$, with $N \trianglelefteq HN$.
2. $H \cap N \trianglelefteq H$.
3. There is a group isomorphism

This is sometimes called the diamond isomorphism theorem because of the lattice shape formed by $HN$, $H$, $N$, and $H \cap N$.

Visualization

Explicit example in $G = \mathbb{Z}$ (additive):

Let $H = 2\mathbb{Z}$ (even integers), $N = 6\mathbb{Z}$ (multiples of 6).

Then $H + N = 2\mathbb{Z}$ (since $6\mathbb{Z} \subset 2\mathbb{Z}$, adding gives $2\mathbb{Z}$) and $H \cap N = \mathrm{lcm}(2,6)\mathbb{Z} = 6\mathbb{Z}$.

       2Z  (= HN)
      /    \
    2Z      6Z
      \    /
       6Z  (= H ∩ N)

Cosets of $6\mathbb{Z}$ in $2\mathbb{Z}$: $\{6\mathbb{Z},\ 2+6\mathbb{Z},\ 4+6\mathbb{Z}\}$ — three cosets. Cosets of $H \cap N = 6\mathbb{Z}$ in $H = 2\mathbb{Z}$: same three cosets.

The isomorphism $2\mathbb{Z}/6\mathbb{Z} \cong 2\mathbb{Z}/6\mathbb{Z}$ is the identity here, but for a non-contained $H$ the identification is nontrivial.

Second example: $G = S_4$, $H = \langle (12) \rangle \cong \mathbb{Z}/2$, $N = V_4 = \{e,(12)(34),(13)(24),(14)(23)\}$.

• $HN = \langle (12) \rangle \cdot V_4$: has order $|H||N|/|H \cap N| = 2 \cdot 4 / 1 = 8$.
• $H \cap N = \{e\}$ (since $(12) \notin V_4$).
• Isomorphism: $HN/N \cong \mathbb{Z}/2/\{e\} \cong \mathbb{Z}/2$.

Coset of $N$ in $HN$	Elements
$N$	$e, (12)(34), (13)(24), (14)(23)$
$(12)N$	$(12), (34), (1324), (1423)$

$|HN/N| = 2 = |H/(H \cap N)|$.

Proof Sketch

1. Define the map $\phi : H \to HN/N$ by $\phi(h) = hN$.
2. $\phi$ is a group homomorphism (inherits from the quotient map $G \to G/N$).
3. $\phi$ is surjective: any coset $hnN = hN$ has preimage $h \in H$.
4. $\ker(\phi) = \{h \in H \mid hN = N\} = \{h \in H \mid h \in N\} = H \cap N$.
5. By the First Isomorphism TheoremFirst Isomorphism Theorem$$G / \ker(f) \cong \operatorname{im}(f)$$The image of a homomorphism is isomorphic to the domain modulo the kernelRead more →, $H/\ker(\phi) \cong \mathrm{im}(\phi) = HN/N$.

Connections

The proof is a one-line application of the First Isomorphism TheoremFirst Isomorphism Theorem$$G / \ker(f) \cong \operatorname{im}(f)$$The image of a homomorphism is isomorphic to the domain modulo the kernelRead more →. The same diamond pattern appears in the proof of the Jordan-Hölder TheoremJordan-Hölder Theorem$$\{0\} \subset H_1 \subset \cdots \subset G \text{ and } \{0\} \subset K_1 \subset \cdots \subset G \text{ give the same factors}$$Any two composition series of a group have the same length and isomorphic composition factors (in some order).Read more →, where it is used to interchange adjacent factors of two composition series. The Third Isomorphism Theorem ($(G/N)/(H/N) \cong G/H$) is the other sibling, completing the trio.

/-- Noether's Second Isomorphism Theorem: given a normal subgroup N ⊴ G and
    any subgroup H ≤ G, there is an isomorphism H/(H ∩ N) ≅ HN/N.
    Mathlib: `QuotientGroup.quotientInfEquivProdNormalQuotient`. -/
theorem second_isomorphism_theorem
    {G : Type*} [Group G] (H N : Subgroup G) [N.Normal] :
    H ⧸ N.subgroupOf H ≃* (H ⊔ N : Subgroup G) ⧸ N.subgroupOf (H ⊔ N) :=
  QuotientGroup.quotientInfEquivProdNormalQuotient H N