First isomorphism theorem

Name
This result is termed the first isomorphism theorem, or sometimes the fundamental theorem of homomorphisms.

General version
Let $$G$$ be a group and $$\varphi:G \to H$$ be a homomorphism of groups. The first isomorphism theorem states that the kernel of $$\varphi$$ is a fact about::normal subgroup, say $$N$$, and there is a natural isomorphism:

$$G/N \cong \varphi(G)$$

where $$\varphi(G)$$ denotes the image in $$H$$ of $$G$$ under $$\varphi$$.

More explicitly, if $$\alpha:G \to G/N$$ is the quotient map, then there is a unique isomorphism $$\psi:G/N \to \varphi(G)$$ such that $$\psi \circ \alpha = \varphi$$.

Version for surjective homomorphism
This is a special case of the more general statement.

Let $$G$$ be a group and $$\varphi:G \to H$$ be a surjective homomorphism of groups. Then, if $$N$$ is the kernel of $$\varphi$$, we have:

$$G/N \cong H$$

More explicitly, if $$\alpha:G \to G/N$$ is the quotient map, then there is a unique isomorphism $$\psi:G/N \to \varphi(G)$$ such that $$\psi \circ \alpha = \varphi$$.

Universal algebraic statement

 * In the variety of groups, every ideal (normal subgroup) is a kernel
 * In the variety of groups, a congruence is completely determined by its kernel. In other words, simply knowing the inverse image of the identity element for a surjective homomorphism, determines the nature of the homomorphism.

This is encoded by saying that the variety of groups is ideal-determined.

Related facts about groups

 * Normal subgroup equals kernel of homomorphism
 * Second isomorphism theorem
 * Third isomorphism theorem
 * Fourth isomorphism theorem

Facts used

 * 1) uses::Normal subgroup equals kernel of homomorphism: Given any homomorphism $$\varphi:G \to H$$ of groups, the kernel of $$\varphi$$ (i.e., the inverse image of the identity element) is a normal subgroup of $$G$$. Further, given any normal subgroup $$N$$ of $$G$$, there is a natural quotient group $$G/N$$.

Proof
Given: A homomorphism of groups $$\varphi:G \to H$$, with kernel $$N$$ (i.e. $$N$$ is the inverse image of the identity element).

To prove: $$N$$ is a normal subgroup, and $$G/N \cong \varphi(G)$$

Proof: Two steps of the proof are done at normal subgroup equals kernel of homomorphism (fact (1)):


 * 1) The kernel of any homomorphism is a normal subgroup
 * 2) If $$N$$ is a normal subgroup, we can define a quotient group $$G/N$$ which is the set of cosets of $$N$$, with multiplication of cosets given by:

$$(aN)(bN) = (ab)N$$

It now remains to show that we can identify $$G/N$$ isomorphically with $$\varphi(G)$$. Consider the map from $$G/N$$ to $$\varphi(G)$$:

$$\overline{\varphi}: aN \mapsto \varphi(a)$$

We first argue that this map is well-defined. For this, observe that if $$b = an, n \in N$$, then:

$$\varphi(b) = \varphi(an) = \varphi(a)\varphi(n) = \varphi(a)$$

In other words, any two elements in the same coset of $$N$$ in $$G$$ get mapped to the same element of $$H$$.

Next, we argue that the map is a group homomorphism. Indeed, if $$aN$$ and $$bN$$ are two cosets, then:

$$\varphi(ab) = \varphi(a)\varphi(b) \implies \overline{\varphi}(abN) = \overline{\varphi}(aN)\overline{\varphi}(bN)$$

(similar checks work for identity element and inverses).

Next, we argue that the map is injective. Indeed, if $$aN$$ is sent to the identity element, then $$\varphi(a) = e$$, forcing $$a \in N$$.

Finally, we argue that the map is surjective. By definition, any element in $$\varphi(G)$$ can be written as $$\varphi(a)$$ for some $$a \in G$$, and hence occurs as $$\overline{\varphi}(aN)$$.

Textbook references

 * , Page 68-69, Theorem (10.9)
 * , Page 97 (Theorem 16): the proof is spread across previous sections, and is not given after the statement