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

\[ A/(H\cap A) \cong H A / H \]

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

\[ H A = H \cup H a_1 \cup H a_2 \cup ... \]

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

\[ H A/H = \{H, H a_1 , H a_2 ,...\} \]

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

\[ A/H = \{H, H a_1 ,H a_2,...\} \]

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

\[ A' = \{H, H a_1 ,H a_2,...\} \]

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

\[ A = H \cup H a_1 \cup H a_2 \cup ... \]

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

\[ A' = \{ H, H a_1 , H a_2 , ... \} \]

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 $x \in G, a \in A$,

\[H x H a H x^{-1} = H a'\]

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.