Suppose $A^{(i)}_{n-1}$ is the alternating group taking the $n-1$ numbers $\{ 1, 2, \cdots, i-1, i+1, \cdots, n \}$ that are the domain of even permutations in it, where $n \geq 4$ and $i =1, 2, \cdots, n$ $($ i.e. $A^{(i)}_{n-1}$ is the set of all even permutation on these above $n-1$ numbers $)$. If fixing index $i$, we can easily get the result that $A^{(i)}_{n-1}$ is a subgroup of $A_n$ for each $i =1, 2, \cdots, n$. Specifically, an element in $A^{(i)}_{n-1}$ can be denoted as the form $$ \left( \begin{array}{ccccccc} 1 & 2 & \cdots & i & \cdots & n-1 & n \\ j_1 & j_2 & \cdots & i & \cdots & j_{n-1} & j_n \end{array} \right), $$
where $j_1,j_2,\cdots,j_n$ is an even permutation of $1,2,\cdots,i-1,i+1,\cdots,n$. Hence, it must be an element in $A_n$.
My question is: how to prove the following statement:
The alternating group $A_n$ is generated by its subgroups $A^{(i)}_{n-1}$, for $i = 1, 2, \cdots, n$.
If $n\ge 5$,for all $\tau=(i,j)(r,s)(i\not=j,r\not=s)$, we prove that this is in the subgroup generated by all $A_{n-1}^i$. This is because if $i=r,j=s$, $\tau=1$ is trivial, if $j=r,i\not=s$, $\tau=(jsi)=(i_1,i_2)(j,s)(s,i)(i_1,i_2)$, here $(i_1,i_2)(j,s)\in A^i_{n-1}$ and $(s,i)(i_1,i_2)\in A^j_{n-1}$($i_1,i_2$ are distinct and not the same as $i,j,s$) . Since each of $A_n^i$ is in $A_n$, the inclusion is thus proved. When $i,j,r,s$ are all distinct, we have $(i,j)(r,s)\in A_{n-1}^{i_1}$. The obstacle remain is to prove the case $n=3$ and $n=4$.
For $n=3$, the theorem does not hold since the only even permutation of two element is $1$.(In your example, I think that $(2,3)$ is not an even permutation, and $(123)\not\in$ any of $A_{n-1}^i$.)
For $n=4$, even permutations for $A_3^i$ are all triples, and triples generates $A_4$