Third Isomorphism Theorem

Groups
Let $$G$$ be a group, and let:


 * $$H, N$$ be normal subgroups of $$G$$
 * $$N$$ be a subset of $$H$$.

Then:
 * $$H / N$$ is a normal subgroup of $$G / N$$
 * where $$H / N$$ denotes the quotient group of $$H$$ by $$N$$


 * $$\frac {G / N} {H / N} \cong \frac G H$$
 * where $$\cong$$ denotes group isomorphism.

Rings
Let $$R$$ be a ring, and let:


 * $$J, K$$ be ideals of $$R$$
 * $$J$$ be a subset of $$K$$.

Then:
 * $$K / J$$ is an ideal of $$R / J$$
 * where $$K / J$$ denotes the quotient ring of $$K$$ by $$J$$;


 * $$\frac {R / J} {K / J} \cong \frac R K$$
 * where $$\cong$$ denotes ring isomorphism.

This result is also referred to by some sources as the first isomorphism theorem, and by others as the second isomorphism theorem.

Proof for Groups

 * We define a mapping $$\phi: G / N \to G / H$$ by $$\phi \left({g N}\right) = g H$$.

Since $$\phi \ $$ is defined on cosets, we need to check that $$\phi \ $$ is well-defined.

Suppose $$x N = y N \implies y^{-1} x \in N$$.

Then $$N \le H \implies y^{-1} x \in H$$ and so $$x H = y H \ $$.

So $$\phi \left({x N}\right) = \phi \left({y N}\right)$$ and $$\phi \ $$ is indeed well-defined.


 * Now $$\phi$$ is a homomorphism, from:

$$ $$ $$


 * Also, since $$N \subseteq H$$, it follows that $$\left|{N}\right| \le \left|{H}\right| \ $$ and so $$\left|{G / N}\right| \ge \left|{G / H}\right| \ $$, indicating $$\phi \ $$ is surjective, so:

$$ $$ $$ $$

The result follows from the First Isomorphism Theorem.

Proof for Rings
In Ring Homomorphism whose Kernel contains Ideal‎, take $$\phi: R \to R / K$$ to be the natural epimorphism.

Then (from the same source) its kernel is $$K$$.

Thus we have that:
 * $$\phi = \psi \circ \nu$$

where $$\psi : R / J \to R / K$$ is a homomorphism.


 * CommDiagThirdIsomTheorem.png

As $$\phi$$ is an epimorphism then from Surjection if Composite is a Surjection we have that $$\psi$$ is a surjection.

So $$\operatorname{Im} \left({\psi}\right) = \operatorname{Im} \left({\phi}\right) = R / K$$ and the First Isomorphism Theorem applies.

Alternative Names
There is no standard numbering for the Isomorphism Theorems. Different authors use different labellings.

This particular result, for example, is also known both as the first isomorphism theorem and the second isomorphism theorem, according to source.