# First isomorphism theorem

## Name

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

## Statement

### 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 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.

## Facts used

1. 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 $\overline{\varphi}(aN)$.

## References

### Textbook references

• Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, More info, Page 68-69, Theorem (10.9)
• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Page 97 (Theorem 16): the proof is spread across previous sections, and is not given after the statement