Theorem
Let \( (G,\cdot) \) be a group and \( H \) be a subgroup of \( G \). Let \( N(H) \) and \( C(H) \) be the normalizer and centralizer of \( H \). Then
\[
\frac{N(H)}{C(H)} \backsimeq \text{ a subgroup of } \text{Aut}(H).
\]
Proof
Step 1
Let us define the homomorphism
For each \( g \in N(H) \), define a mapping
\[
\phi(g): H \to H \quad \text{by} \quad \phi(g)(h) = g h g^{-1} \quad \forall\, h \in H.
\]
Since \( g \in N(H) \), we have \( gHg^{-1} = H \); hence, \( \phi(g) \) is well-defined and is an automorphism of \( H \).
Step 2
To prove \( \phi \) is a Homomorphism
For \( g_1, g_2 \in N(H) \) and for all \( h \in H \):
\begin{align}
\phi(g_1g_2)(h) &= (g_1g_2) \, h \, (g_1g_2)^{-1} \nonumber\\
&= g_1 \Bigl( g_2 \, h \, g_2^{-1} \Bigr) g_1^{-1} \nonumber\\
&= \phi(g_1)\bigl(\phi(g_2)(h)\bigr) \nonumber
\end{align}
Thus, \( \phi(g_1g_2) = \phi(g_1) \circ \phi(g_2) \); that is, \( \phi \) is a group homomorphism from \( N(H) \) to \( \text{Aut}(H) \).
Step 3
To find the Kernel of \( \phi \)
The kernel of \( \phi \) is the set of all elements \( g \in N(H) \) such that \( \phi(g) \) is the identity automorphism on \( H \):
\[
\ker \phi = \{ g \in N(H) : g h g^{-1} = h \ \text{for all } h \in H \}.
\]
This set is precisely the centralizer \( C(H) \). Hence,
\begin{align}
\ker \phi = C(H)\nonumber
\end{align}
Step 4
Let us apply the First Isomorphism Theorem
By the First Isomorphism Theorem, since \( \phi: N(H) \to \text{Aut}(H) \) is a homomorphism with kernel \( C(H) \), we have:
\begin{align}
\frac{N(H)}{C(H)} \cong \phi(N(H)) \nonumber
\end{align}
Here, \( \phi(N(H)) \) is a subgroup of \( \text{Aut}(H) \), which completes the proof.
