Third isomorphism theorem

Statement
Suppose $$G$$ is a group, and $$H$$ and $$K$$ are fact about::normal subgroups of $$G$$, such that $$H \le K$$. Then we have the following natural isomorphism:

$$(G/H)/(K/H) \cong G/K$$

Where the isomorphism sends a coset $$Hg$$ in $$G$$ to the coset $$Kg$$ in $$G$$.

Note that this statement makes sense at the level of a group isomorphism only when both $$H$$ and $$K$$ are normal in $$G$$. Otherwise, the statement is still true at the level of sets, but we cannot make sense of it as a group isomorphism.

Other isomorphism theorems

 * First isomorphism theorem
 * Second isomorphism theorem
 * Fourth isomorphism theorem (also known as the lattice isomorphism theorem or the correspondence theorem)
 * Zassenhaus isomorphism theorem

Facts used

 * Normality satisfies intermediate subgroup condition
 * Normality is image-closed (not really required; implicitly shown in the proof)
 * First isomorphism theorem

Proof
Given: A group $$G$$, with normal subgroups $$H$$ and $$K$$, such that $$H \le K$$

To prove: $$(G/H)/(K/H) \cong G/K$$

Proof: Note first that all the three expressions for quotient groups make sense. $$G/H$$ and $$G/K$$ make sense because $$H,K$$ are normal in $$G$$. Moreover, since normality satisfies intermediate subgroup condition, $$H$$ is also normal in $$K$$.

Next, observe that $$K/H$$ is a normal subgroup in $$G/H$$, because normality is image-closed: under the quotient map by $$H$$, the normal subgroup $$K$$ of $$G$$ gets sent to a normal subgroup $$K/H$$ of $$G/H$$. Thus, the left side makes sense.

Let's now describe the isomorphism from the left side to the right side:

$$\psi: G/H \to G/K, \qquad \psi(gH) = gK$$

In other words, the map takes a coset of $$H$$ and gives the corresponding coset of $$K$$. This is well-defined, because if $$h \in H$$, then $$h \in K$$, so $$(gh)K = g(hK) = gK$$.

Further, the map is a homomorphism. For this, observe that it sends the identity element to the identity element, preserves the group multiplication, and preserves the inverse map.

Further, the map is surjective, because any coset $$gK$$ occurs as the image of $$gH$$ under $$\psi$$.

Finally, we need to determine the kernel of the map. This is given by the set of $$gH$$ such that $$gK = K$$. This is precisely those cosets of $$H$$ that are in $$K$$, which is the same as the coset space $$K/H$$. Hence, the kernel of the map is precisely $$G/K$$.

Thus, the surjective homomorphism $$\psi:G/H \to G/K$$ has kernel precisely $$K/H$$. By the first isomorphism theorem, we get:

$$(G/H)/(K/H) \cong G/K$$.

Textbook references

 * , Theorem 19, Section 3,3, Page 98
 * , Exercise 8, Miscellaneous Problems, Page 236