View source for Third isomorphism theorem

{{basic fact}}
{{isomorphism theorem}}
[[difficulty level::2| ]]
==Statement==

Suppose <math>G</math> is a [[group]], and <math>H</math> and <math>K</math> are [[fact about::normal subgroup]]s of <math>G</math>, such that <math>H \le K</math>. Then we have the following natural isomorphism:

<math>(G/H)/(K/H) \cong G/K</math>

Where the isomorphism sends a coset <math>Hg</math> in <math>G</math> to the coset <math>Kg</math> in <math>G</math>.

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

==Related facts==

===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 <math>G</math>, with normal subgroups <math>H</math> and <math>K</math>, such that <math>H \le K</math>

''To prove'': <math>(G/H)/(K/H) \cong G/K</math>

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

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

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

<math>\psi: G/H \to G/K, \qquad \psi(gH) = gK</math>

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

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 <math>gK</math> occurs as the image of <math>gH</math> under <math>\psi</math>.

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

Thus, the surjective homomorphism <math>\psi:G/H \to G/K</math> has kernel precisely <math>K/H</math>. By the [[first isomorphism theorem]], we get:

<math>(G/H)/(K/H) \cong G/K</math>.
==References==

===Textbook references===
* {{booklink-proved|DummitFoote}}, Theorem 19, Section 3,3, Page 98
* {{booklink-stated|Artin}}, Exercise 8, Miscellaneous Problems, Page 236