The isomorphism theorems describe important relationships among homomorphisms, normal subgroups, quotient groups, and images of groups.


First Isomorphism Theorem

Theorem: Let

\[\psi:G\rightarrow G'\]

be a group homomorphism. Then \(\ker\psi\) is a normal subgroup of \(G\), and

\[\frac{G}{\ker\psi}\cong\operatorname{Im}\psi.\]

In particular, if \(\psi\) is surjective, then \(\operatorname{Im}\psi=G'\), and hence

\[\frac{G}{\ker\psi}\cong G'.\]

The isomorphism is given by

\[x\ker\psi\longmapsto\psi(x).\]

This theorem states that the image of a homomorphism is isomorphic to the domain group after the kernel has been factored out.


Second Isomorphism Theorem

Theorem: Let \(H\) and \(K\) be subgroups of a group \(G\), and suppose that \(K\) is normal in \(G\). Then \(H\cap K\) is normal in \(H\), \(HK\) is a subgroup of \(G\), and

\[\frac{H}{H\cap K}\cong\frac{HK}{K}.\]

The corresponding isomorphism is induced by the homomorphism

\[\phi:H\rightarrow\frac{HK}{K},\qquad \phi(h)=hK.\]

Its kernel is \(H\cap K\), and its image is \(HK/K\). Therefore, the result follows from the First Isomorphism Theorem.


Third Isomorphism Theorem

Theorem: Let \(H\) and \(K\) be normal subgroups of a group \(G\) such that

\[H\subseteq K.\]

Then \(K/H\) is a normal subgroup of \(G/H\), and

\[\frac{G/H}{K/H}\cong\frac{G}{K}.\]

This theorem is sometimes called the quotient of a quotient theorem. It shows that first factoring \(G\) by \(H\) and then factoring the resulting group by \(K/H\) gives, up to isomorphism, the same result as factoring \(G\) directly by \(K\).


Maximal Normal Subgroup

Let \(G\) be a group. A normal subgroup \(N\) of \(G\) is called a maximal normal subgroup of \(G\) if the following conditions hold:

Thus, a maximal normal subgroup is a proper normal subgroup that is not properly contained in any other proper normal subgroup of \(G\).


Simple Group

A nontrivial group \(G\) is called a simple group if its only normal subgroups are

\[\{e\}\quad\text{and}\quad G.\]

Equivalently, a nontrivial group is simple if it has no nontrivial proper normal subgroup.


Maximal Normal Subgroups and Simple Quotients

Corollary: Let \(N\) be a proper normal subgroup of \(G\). Then \(N\) is a maximal normal subgroup of \(G\) if and only if the quotient group \(G/N\) is simple.

This result provides a useful criterion for determining whether a normal subgroup is maximal.


Intersection of Distinct Maximal Normal Subgroups

Corollary: Let \(H\) and \(K\) be distinct maximal normal subgroups of \(G\). Then:

Indeed, by the Second Isomorphism Theorem,

\[\frac{H}{H\cap K}\cong\frac{G}{K}\]

and

\[\frac{K}{H\cap K}\cong\frac{G}{H}.\]

Since \(G/K\) and \(G/H\) are simple, \(H\cap K\) is maximal normal in both \(H\) and \(K\).