Second Isomorphism Theorem

Theorem
Let $$H \le G, N \triangleleft G$$.

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

Proof

 * 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.

Comment
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.