Prove that $S_4$ cannot be generated by $(1 3),(1234)$ Prove that $S_4$ cannot be generated by $(1 3),(1234)$
I have checked some combinations between $(13),(1234)$ and found out that those combinations cannot generated 3-cycles.
Updated idea:
Let $A=\{\{1,3\},\{2,4\}\}$
Note that $(13)A=A,(1234)A=A$
Hence, $\sigma A=A,\forall\sigma\in \langle(13),(1234)\rangle$
In particular, $(12)\notin \sigma A,\forall\sigma\in \langle(13),(1234)\rangle$
So we conclude that $S_4\neq\langle(13),(1234)\rangle$
 A: The partition $\{\{1,3\},\{2,4\}\}$ is invariant under the action of the two proposed generators, but not under all of $S_4$, so they cannot generate all of $S_4$.
A: If we denote $a = (1234)$ and $b = (13)$, one can easily check that $a^4 = e$, $b^2 = e$ and $ab = ba^{-1}$ which are precisely relations that define dihedral group $D_4$. Thus, subgroup generated by $a$ and $b$ in $S_4$ is isomorphic to quotient of $D_4$, and thus it's order is less or equal than $8$. Since $|S_4| = 4!$, obviously $a$ and $b$ can't generate whole $S_4$. One can easily check that there are $8$ distinct elements in $\langle a,b\rangle$, so it is actually isomorphic to $D_4$.
A: Here we copy/paste/modify another answer:
Define
$\; \tau = (13)$
$\;\sigma = (1234)$


$1^{st} \text {group of calculations:}$
$\;\tau^1 = (13)$
$\;\tau^2 = \tau^0 \quad \text { - the identity permutation}$


$2^{nd} \text{ group of calculations:}$
$\;\sigma^1 = (12)\,(23) \,(34)$
$\;\sigma^2 = (13)\,(24)$
$\;\sigma^3 = (14)\,(24)\,(34)$
$\;\sigma^4 = \sigma^0 \quad \text { - the identity permutation}$


$3^{rd} \text{ group of calculations:}$
$\tau\sigma = (12) \,(34)$
$\tau\sigma^2 = (24)$
$\tau\sigma^3 = (14) \,(23)$


$4^{th} \text{ group of calculations:}$
$\sigma\tau = (14) \,(23)$
$\sigma^2\tau = (24)$
$\sigma^3\tau = (12)\,(34)$



So far we've identified exactly $8$ permutations that are in the group generated by $\tau$ and $\sigma$.
When we run the $4^{th} \text{ group of calculations}$ we get the same permutations as the $3^{rd} \text{ group of calculations}$, and we can now write these symbolic (defining) rules,
$\tag 1 \tau^2 = 1_d$
$\tag 2 \sigma^4 = 1_d$
$\tag 3 \sigma\tau = \tau\sigma^3$
Given any word (string) in the letters $\tau$ and $\sigma$ we can 'move' all the $\tau$ letters to the left and standardize (present/represent) the permutation to have the form
$\tag 4 \tau^n \sigma^m \quad n \in \{0,1\} \text{ and } m \in \{0,1,2,3\}$
We conclude that the group generated by $\tau$ and $\sigma$ has exactly $8$ elements.
Since
$\; \sigma \tau = \tau \sigma^3$
and
$\; \tau \sigma \ne \tau \sigma^3$
we also know that this a non-abelian subgroup of $S_4$.
