Relevant Facts
Proposition $1$: Suppose $G$ is a group with subgroups $H,K < G$ such that $H \lhd G$ and $H \cap K = \{e\}$. Then the map: $$\phi: H \rtimes_c K \to HK\ \text{defined by}\ \phi(h,k) = hk$$is an isomorphism. Where $c: K \to Aut(H)$ is conjugation in $G$.
Proposition $2$: For any group homomorphism $\phi: G \to H$, $\text{Ker}(\phi) \lhd G$.
Corollary $1$: If $\sigma \in S_n$ is a $k$-cycle then $\sigma$ is even if $k$ is odd and vice versa.
For this problem, I approached it by following Proposition $1$ and taking $H = A_n$ (where $A_n$ is the alternating group of order $n$ which is a subgroup of $S_n$ that contains all even permutations) and trying to find a subgroup $K < S_n$ such that $K \cong \mathbb{Z}_2$. To that end:
My guess is to take $K$ to be one of the cyclic subgroups generated by a transposition. So take $K = \langle (12) \rangle = \{(1),(12)\}$. Then we can immediately see that since $A_n$ consists of only even permutations and $\langle (12) \rangle = \{(1),(12)\}$ where $(12)$ is an odd permutation by Corollary $1$ that $A_n \cap \langle (12) \rangle = \{(1)\}$. Additionally $A_n$ is the kernel of the sign homomorphism $\epsilon: S_n \to C_2 = \{ \pm 1\}$ and so by proposition $2$, $A_n \lhd S_n$. Now, we want to show that $\langle (12) \rangle \cong \mathbb{Z}_2$. To that end we have the mapping $\lambda: \langle (12) \rangle \to \mathbb{Z}_2$ which identifies $$\lambda((1)) = [0]$$ $$\lambda((12)) = [1]$$ Clearly this is a bijection since we can construct the inverse mapping $\lambda^{-1}: \mathbb{Z}_2 \to \langle (12) \rangle$ by identifying: $$\lambda^{-1}([0]) = (1)$$ $$\lambda^{-1}([1]) = (12)$$ This is a homomorphism as well since for $k,k' \in \langle (12) \rangle$ and $[z],[z'] \in \mathbb{Z}_2$ we have $$\phi(kk') = [z + z'] = [z] + [z'] = \phi(k)\phi(k')$$ Thus by Proposition $1$ we can see that $A_n \rtimes \mathbb{Z}_2 \cong A_n\mathbb{Z}_2$. Now all that remains is to show that $A_n\mathbb{Z}_2 = S_n$
Up to this point I feel confident in my proof but would like confirmation that I'm following the correct line of thinking and my arguments are sound. Additionally, is my proof of $\lambda$ as an isomorphism correct? In particular, can I just simply construct the mappings since we are dealing with $2$ finite groups? Then declare the inverse mapping in the way that I have?
I would also appreciate any help in proving that $A_n\mathbb{Z}_2 = S_n$. My guess would be to show that since $[0]$ is isomorphic to $(1)$ then for any $a \in A_n$ the product $a[0] \in A_n\mathbb{Z}_n$ can be evaluated as $a(1) = a$. Thus $A_n(1) = A_n$ and so $A_n\mathbb{Z}_2$ consists of all even permutations and the identity. Then I was hoping to show that $A_n[1] \in A_n\mathbb{Z}_2$ would yield all odd permutations. To that end I said suppose $a \in A_n$ is any $k$-cycle where $k = 2q +1 (q \in \mathbb{Z})$ is odd. Then since $(12)$ is a $2$-cycle and is isomorphic to $[1]$ the product $a[1] \in A_n\mathbb{Z}_2$ can be evaluated as $a(12)$ which is a $k+2$-cycle, but this still yields an odd number and hence an even permutation.
It is trivial that $\{id,(1\ 2)\}$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ simply because $(1\ 2)$ is of order $2$. You already proved that $\mathfrak{A}_n\triangleleft\mathfrak{S}_n$ and $\mathfrak{A}_n\cap\langle(1\ 2)\rangle=\{id\}$. Now the application $\mathfrak{A}_n\times\langle(1\ 2)\rangle\longrightarrow\mathfrak{S}_n$ defined by $(\sigma,k)\mapsto (1\ 2)^k\sigma$ is NOT a morphism, but it is injective (take the parity of the permutation), since the two sets have the same cardinality, this is a surjection therefore $\mathfrak{S}_n=\mathfrak{A}_n\cdot\langle(1\ 2)\rangle$.