Second isomorphism theorem

Name
This result is termed the second isomorphism theorem or the diamond isomorphism theorem (the latter name arises because of the diamond-like shape that can be used to describe the theorem).

General statement
Suppose $$G$$ is a group, and $$H,N$$ are two subgroups of $$G$$ such that $$H$$ normalizes $$N$$; in other words $$hN = Nh$$ for every $$h \in H$$. Then, $$HN$$ is a group, with $$N \triangleleft HN$$ (i.e., $$N$$ is a fact about::normal subgroup of $$HN$$), and $$H \cap N \triangleleft H$$ (i.e. $$H \cap N$$ is a normal subgroup of $$H$$), and:

$$HN/N \cong H/(H \cap N)$$

Particular case
Suppose $$G$$ is a group, $$N$$ is a normal subgroup, and $$H$$ an arbitrary subgroup, such that $$HN = G$$. Then:

$$G/N \cong H/(H \cap N)$$



Specific possibilities for the relationship between $$H$$ and $$N$$
Note that for each of these, if we have $$HN = G$$, then the statement made for $$HN$$ applies to $$G$$.

Facts about normal subgroups

 * Normality satisfies transfer condition: If $$H, N \le G$$ are subgroups and $$N$$ is normal in $$G$$, then $$H \cap N$$ is normal in $$H$$.
 * Normality satisfies intermediate subgroup condition: If $$H \le K \le G$$ are groups and $$H$$ is normal in $$G$$, then $$H$$ is normal in $$K$$.

General version of the result

 * Product formula: The set-theoretic version of the product formula establishes a bijection which is the same as the bijection of the second isomorphism theorem, but without the conditions of normality. The bijection is purely at the set-theoretic level.

Facts used

 * 1) uses::Subgroup containment implies coset containment

Proof
Given: A group $$G$$, subgroups $$H, N \le G$$ such that $$hN = Nh$$ for all $$h \in H$$.

To prove: $$HN$$ is a group, $$N$$ is normal in $$HN$$, $$H \cap N$$ is normal in $$H$$, and $$HN/N \cong H/(H \cap N)$$.

Proof:

$$N$$ is normal in $$HN$$
Observe that, by the condition, $$hNh^{-1} = N$$ for any $$h \in H$$. Thus, for $$h \in H$$ and $$n \in N$$, we have $$hnN(hn)^{-1} = h(nNn^{-1})h^{-1} = hNh^{-1} = N$$. Thus, $$N$$ is normal in $$HN$$.

$$H \cap N$$ is normal in $$H$$
We now prove that $$H \cap N$$ is normal in $$H$$. Pick $$h \in H, x \in H \cap N$$. Then $$hxh^{-1} \in H$$ such $$H$$ is a subgroup. Also, since $$hNh^{-1} =N$$ for any $$h \in H$$, we have $$hxh^{-1} \in N$$, so $$hxh^{-1} \in H \cap N$$. Thus, $$h(H \cap N)h^{-1} \subseteq H \cap N$$, and we get that $$H \cap N$$ is normal in $$H$$.

Definition of isomorphism and proof that it works
Finally, define the isomorphism $$\varphi$$ between $$H/(H \cap N)$$ and $$\! HN/N$$ as follows:

$$\varphi(g(H \cap N)) = gN$$.

We check that this satisfies all the required conditions: