## The Isomorphism Theorems

Let $G$ be a group. Let $H \triangleleft G$. Then a *natural
homomorphism* exists from $G$ to $G/H$, given by
$g \mapsto H g$.

**First Isomorphism Theorem**:
Let $\phi : G \rightarrow G'$ be a group homomorphism.
Let $E$ be the subset of $G$ that is mapped to the identity of $G'$.
$E$ is called the *kernel* of the map $\phi$.
Then $E\triangleleft G$ and $G/E \cong im \phi$.

An *automorphism* is an isomorphism from a group $G$ to itself.
Let $g \in G$. Then the map that sends $a\in G$ to $g^{-1} a g$ is an
automorphism. Automorphisms of this form are called *inner*
automorphisms, otherwise they are called *outer*
automorphisms. Note that all inner automorphisms of an abelian group
reduce to the identity map.

**Second Isomorphism Theorem:** Let $G$ be a group. Let $H\triangleleft G$.
If $A$ is any subgroup of $G$, then $H\cap A$ is normal and

**Proof:** Let $A = \{1, a_1,a_2,..\}$. Then the image of $A$ under the
natural map from $G$ to $G/H$ is

Now $H A$ is a subgroup by the Product Theorem because $H A = A H$ since $H$ is normal, and $H$ is normal in $H A$, thus

Lastly, the kernel of the natural map from $G$ to $G/H$ when restricted to $A$ is clearly $H \cap A$, and applying the first isomorphism theorem proves the result.

**Theorem:**
If $H\triangleleft G$ and $A$ is some subgroup satisfying
$H\subset A\subset G$ then
$A/H$ is a subgroup of $G/H$.
Conversely, every subgroup of $G/H$ is of the form $A/H$ for some
$H\subset A \subset G$.

**Proof:**
$H$ is normal in $G$, so $H$ must also be normal in $A$. Let
$A = \{1,a_1,a_2...\}$. Then

must be some subgroup of $G/H$. Conversely, suppose

is some subgroup of $G/H$. Then the set

is a subgroup of $G$ since for any $h_1, h_2\in H$ we have $h_1 a_i h_2 a_j = h_3 a_k$ where $a_k = a_i a_j$ and for some $h_3\in H$, since $A'$ is a group. Thus $A' = A/H$.

**Third Isomorphism Theorem:**
If $H\triangleleft G$ and $H\triangleleft A\triangleleft G$ then
$A/H \triangleleft G/H$ and $\frac{G/H}{A/H} \cong G/A$.
Conversely, every normal subgroup of $G/H$ is of the form $A/H$ for some
$H\triangleleft A \triangleleft G$.

**Proof:**
Consider the map from $G/H \rightarrow G/A$ that sends $H x$ to $A x$.
The map is well-defined because $H x = H y$ implies $x y^{-1} \in H \subset A$
whence $A x = A y$. This map is homomorphic because $H x H y = H x y$ is
mapped to $A x y = A x A y$. The kernel of the map consists of all elements
of $G/H$ that get mapped to $A$, in other words, elements of the form $H x$
with $A x = A$. This happens if and only if $x \in A$, thus the the kernel
consists of the cosets of the form $H a$ for $a\in A$. That is, the
kernel is precisely $A/H$. By the first isomorphism theorem, $A/H$ is
therefore normal in $G/H$ and we have
$\frac{G/H}{A/H} \cong G/A$.

Conversely, suppose

is a normal subgroup of $G/H$. Then we know that $A = H \cup H a_1 \cup H a_2 \cup ...$ is a subgroup of $G$, and it remains to show $A$ is normal. Since $A'$ is normal, we have for all $x \in G, a \in A$,

for some $a' \in A$. In particular, by picking the identity for the first and third occurences of $H$ in the equation, $x H a x^{-1} \subset H a'$ for some $a' \in A$, and hence $x A x^{-1} \subset H \cup H a_1 \cup H a_2 \cup ... \subset A$. Swapping $x$ with its inverse gives the reverse inclusion $x A x^{-1} \supset A$, thus $A = x^{-1} A x$, that is, $A$ is normal.