Second Isomorphism Theorem

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


 * $$H$$ be a subgroup of $$G$$.
 * $$N$$ be a normal subgroup of $$G$$.

Then:
 * $$\frac H {H \cap N} \cong \frac {H N} N$$

where $$\cong$$ denotes group isomorphism.

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


 * $$S$$ be a subring of $$R$$.
 * $$J$$ be an ideal of $$R$$.

Then:


 * $$(1) \quad S + J$$ is a subring of $$R$$
 * $$(2) \quad J$$ is an ideal of $$S + J$$
 * $$(3) \quad S \cap J$$ is an ideal of $$S$$
 * $$(4) \quad \frac S {S \cap J} \cong \frac {S + J} J$$

where $$\cong$$ denotes group isomorphism.

This result is also referred to by some sources as the first isomorphism theorem.

Proof for Groups

 * From Quotient Group, for $$G / H$$ to be defined, it is necessary for $$H \triangleleft G$$.

The fact that Intersection with Normal Subgroup is Normal gives us that $$N \cap H \triangleleft H$$.

Also, $$N \triangleleft N H = \left \langle {H, N} \right \rangle$$ follows from Subgroup Product with Normal Subgroup as Generator.


 * Now we define a mapping $$\phi: H \to H N / N$$ by the rule $$\phi \left({h}\right) = h N$$.

Note that $$N$$ need not be a subset of $$H$$. Therefore, the coset $$h N$$ is an element of $$H N / N$$ rather than of $$H / N$$.

Then $$\phi$$ is a homomorphism, as $$\phi \left({x y}\right) = x y N = \left({x N}\right) \left({y N}\right) = \phi \left({x}\right) \phi \left({y}\right)$$.

Then:

$$ $$ $$ $$

Then we see that $$\phi$$ is a surjection because $$h n N = h N \in H N / N$$ is $$\phi \left({h}\right)$$.

The result follows from the First Isomorphism Theorem.

Proof for Rings
The relations being defined can be illustrated by this commutative diagram:


 * CommDiagSecondIsomTheorem.png


 * $$(1) \quad S + J$$ is a subring of $$R$$

From Sum of All Ring Products is Additive Subgroup, $$S + J$$ is an additive subgroup of $$R$$.

Suppose $$s, s' \in S, j, j' \in J$$.

Then:
 * $$\left({s + j}\right) \left({s' + j'}\right)$$

$$ $$

so by the Subring Test $$S + J$$ is a subring of $$R$$.


 * $$(2) \quad J$$ is an ideal of $$S + J$$

Let $$s + j \in S + J$$ and let $$j \in J$$.

Then:

$$ $$

So $$J$$ is an ideal of $$S + J$$.


 * $$(3) \quad S \cap J$$ is an ideal of $$S$$

Let $$\nu: R \to R / J$$ be the natural epimorphism.

Let $$\nu'$$ be the restriction of $$\nu$$ to $$S$$.

Then $$\nu': S \to R / J$$ is a homomorphism.

The image of $$\nu'$$ is the set of all cosets $$s + J$$ for $$s \in S$$:
 * $$\operatorname{Im} \left({\nu'}\right) = \frac {S + J} J$$

Now, the kernel of $$\nu'$$ is the set of all elements of $$S$$ which are sent to $$0_{S/J}$$ by $$\nu$$.

That is, all the elements of $$S$$ which are also in $$J$$ itself, which is how the quotient ring behaves.

That is:
 * $$\ker \left({\nu'}\right) = S \cap J$$

and so from Kernel of Ring Homomorphism is Ideal, $$S \cap J$$ is an ideal of $$S$$.


 * $$(4) \quad \frac S {S \cap J} \cong \frac {S + J} J$$

This follows directly from the First Isomorphism Theorem.

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

This particular result, for example, is also known as the first isomorphism theorem.