First isomorphism theorem

This article gives the statement, and possibly proof, of a basic fact in group theory.
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This article is about an isomorphism theorem in group theory.
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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 $\phi :G\to H$ be a homomorphism of groups. The first isomorphism theorem states that the kernel of $\phi$ is a Normal subgroup (?), say $N$, and there is a natural isomorphism:

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

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

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

Version for surjective homomorphism

This is a special case of the more general statement.

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

$G/N\cong H$

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

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

Related facts about groups

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

Facts used

1. Normal subgroup equals kernel of homomorphism: Given any homomorphism $\phi :G\to H$ of groups, the kernel of $\phi$ (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 $\phi :G\to H$, with kernel $N$ (i.e. $N$ is the inverse image of the identity element).

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

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 $\phi (G)$. Consider the map from $G/N$ to $\phi (G)$:

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

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

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

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:

$\phi (ab)=\phi (a)\phi (b)⟹\overline{\phi }(abN)=\overline{\phi }(aN)\overline{\phi }(bN)$

(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 $\phi (a)=e$, forcing $a\in N$.

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

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