A splitting lemma for groups

We recall the definition of the semidirect product of two groups. Suppose $N$ and $H$ are two groups and $H$ acts on $N$ by $\vphi$, i.e. we have a homomorphism $\func{\vphi}{H}{\mathrm{Aut}(N)}$, $h \mapsto \vphi_h$. The semidirect product $N \rtimes_\vphi H$ is defined by endowing the set $N \times H$ the multiplication \[ (n_1, h_1) \cdot (n_2, h_2) = (n_1 \vphi_{h_1}(n_2), h_1 h_2). \] Identifying $N$ with $N \times \{1\}$ and $H$ with $\{1\} \times H$, $N$ and $H$ are subgroups of $N \rtimes_\vphi H$ with $N$ normal.

Theorem (Splitting Lemma). Let $G$ be a group, $H$ be a subgroup of $G$ and $N$ be a normal subgroup of $G$. Then $G$ is isomorphic to a semidirect product $N \rtimes_\vphi H$ if and only if
(i) there exist a short exact sequence $1 \rightarrow N \overset{\iota}\rightarrow G \overset{\pi}{\rightarrow} H \rightarrow 1$ and
(ii) a homomophism $\func{\alpha}{H}{G}$ such that $\pi \circ \alpha = \mathrm{id}_H$.
Proof: $(\Leftarrow)$ Suppose $G = N \rtimes_\vphi H$. To prove (i) let $\iota$ be the inclusion map of $N$ into $G$, which is a injective homomorphism. Define $\func{\pi}{G}{H}$ by $\pi((n, h)) = h$. Clearly $\pi$ is a surjective homomorphism and its kernel is $N \times \{1\} \cong N$. For (ii), define $\func{\alpha}{H}{G}$ by $\alpha(h) = (1, h)$. Then $\alpha$ is a homomorphism with $\pi(\alpha(h)) = \pi((1, h)) = h$ for all $h \in H$.
$(\Rightarrow)$ Define a homomorphism $\func{\vphi}{H}{\mathrm{Aut}(N)}$ by $\vphi_h(n) = \iota^{-1}(\alpha(h)\iota(n)\alpha(h)^{-1})$, note that since $N \lhd G$ we have $gng^{-1} \in N$ whenever $g \in G$ and $n \in N$, so $\vphi$ is well-defined. We claim that $G \cong N \rtimes_\vphi H$. Define $\func{\Psi}{N \rtimes_\vphi H}{G}$ by $\Psi((n, h)) = \iota(n)\alpha(h)$. It is a homomorphism since \[ \begin{align*}\Psi((n_1, h_1)\cdot(n_2, h_2)) &= \iota(n_1)\alpha(h_1)\iota(n_2)\alpha(h_1)^{-1}\alpha(h_1 h_2) \\
&=  \iota(n_1)\alpha(h_1)\iota(n_2)\alpha(h_2) \\
&= \Psi((n_1, h_1))\Psi((n_2, h_2)). \end{align*} \] Now define $\func{\Phi}{G}{N \rtimes_\vphi H}$ by $\Phi(g) = (\iota^{-1}(g (\alpha \circ \pi)(g^{-1})), \pi(g))$. It is well-defined since \[ \pi(g(\alpha \circ \pi)(g^{-1})) = \pi(g)(\pi \circ \alpha)(\pi(g)^{-1}) = \pi(g)\pi(g)^{-1} = 1 \] and thus $g(\alpha \circ \pi)(g^{-1}) \in \ker(\pi) = \mathrm{im}(\iota)$. It is a homomorphism as
\[ \begin{align*} \Phi(g_1)\Phi(g_2) &= (\iota^{-1}(g_1 (\alpha \circ \pi)(g_1^{-1})), \pi(g_1)) \cdot  (\iota^{-1}(g_2 (\alpha \circ \pi)(g_2^{-1})), \pi(g_2)) \\
&= (\iota^{-1}(g_1 (\alpha \circ \pi)(g_1^{-1})) \vphi_{\pi(g_1)}(\iota^{-1}(g_2 (\alpha \circ \pi)(g_2^{-1}))), \pi(g_1)\pi(g_2)) \\
&= (\iota^{-1}(g_1 (\alpha \circ \pi)(g_1^{-1})) \iota^{-1}(\alpha(\pi(g_1))g_2(\alpha \circ \pi)(g_2^{-1})\alpha(\pi(g_1))^{-1}), \pi(g_1 g_2)) \\
&= (\iota^{-1}(g_1 g_2 (\alpha \circ \pi)(g_2^{-1} g_1^{-1})), \pi(g_1 g_2)) \\
&= \Phi(g_1 g_2).
\end{align*}\] Now \[ \begin{align*}\Phi \circ \Psi(n, h) &= \Phi(\iota(n)\alpha(h)) = \Phi(\iota(n))\Phi(\alpha(h)) \\
&= (\iota^{-1}(\iota(n)(\alpha \circ \pi \circ \iota)(n^{-1})), (\pi \circ \iota)(n)) \cdot (\iota^{-1}(\alpha(h)(\alpha \circ \pi \circ \alpha)(h^{-1})), (\pi \circ \alpha)(h)) \\
&= (n, 1) \cdot (1, h) = (n, h)
\end{align*} \] and \[ \begin{align*} \Psi \circ \Phi(g) &= \Psi((\iota^{-1}(g (\alpha \circ \pi)(g^{-1})), \pi(g))) \\
&= \iota(\iota^{-1}(g (\alpha \circ \pi)(g^{-1})))\alpha(\pi(g)) = g.
\end{align*} \] Hence $\Phi$ and $\Psi$ are inverse to each other and thus are isomorphisms.

A proof of Cayley-Hamilton Theorem using Zariski topology

The notation $\mathbb{A}^n$ is used to denote the space $\mathbb{C}^n$ with the Zariski topology.

We note the following basic properties of the Zariski topology.

Proposition 1. Nonempty open subsets of $\mathbb{A}^n$ is dense.

Lemma 2. Polynomial maps from $\mathbb{A}^n$ to $\mathbb{A}^m$ is continuous.

We proceed to prove the Cayley-Hamilton Theorem. Identify $M_n(\mathbb{C})$ with $\mathbb{A}^{n^2}$. For any matrix $A$, let $p_A(\lambda)$ denotes its characteristic polynomial. The Cayley-Hamilton Theorem is the statement that the map $\Psi: \mathbb{A}^{n^2} \rightarrow \mathbb{A}^{n^2}$, $\Psi(A) = p_A(A)$ vanishes identically. Let $D$ be the subset of $\mathbb{A}^{n^2}$ of matrices with distinct eigenvalues.

Lemma 3. $D$ is nonempty and open.
Proof: Clearly $D$ is not empty. The complement of $D$ is the collection of n x n matrices with repeated eigenvalues. But a matrix has repeated eigenvalues if and only if the discriminant of its characteristic polynomial vanishes. Hence $D$ is closed.

It follows from Proposition 1 and Lemma 3 that $D$ is dense in $\mathbb{A}^{n^2}$. Note that $\Psi$ is a polynomial map, so it is continuous by Lemma 2. Hence it suffices to show that $\Psi$ vanishes on $D$. Let $A \in D$. As $A$ is diagonalizable, there exists an invertible matrix $P$ such that $P^{-1}AP$ is diagonal. Note that $p_A(A) = p_{P^{-1}AP}(P^{-1}AP)$. So we may assume $A$ itself is diagonal. If $A = \mathrm{diag}(\lambda_1, \ldots, \lambda_n)$, then $p_A(\lambda) = \prod_{i = 1}^n (\lambda - \lambda_i)$, thus it is clear that $p_A(A) = 0$.