$S_4 \ne \langle (1,2,3,4), \, (1,3)\rangle$ So I'm trying to prove $S_4≠⟨(1,2,3,4),(1,3)⟩$, and I get the basic idea that $(1,2)$ swaps two things next to each other, which neither of the other operations do, and necessarily neither do their products. But I'm having a hard time formalizing this argument. Of course, the hard part is arguing that no product of these two could just swap neighbors. I know I could prove this by enumeration but that's boring.
 A: Define a metric $d$ on $\{1,2,3,4\}$ as follows: $d(1,2)=d(2,3)=d(3,4)=d(4,1)=1$;
$d(1,3)=d(2,4)=\sqrt2;\ d(x,y)=d(y,x);\ d(x,x)=0.$
An isometry of the metric space $\{1,2,3,4\}$ is a bijection $f:\{1,2,3,4\}\to\{1,2,3,4\}$ which preserves distances, i.e., $d(f(x),f(y))=d(x,y)$ for all $x,y\in\{1,2,3,4\}$. Observe that the set of all isometries of $\{1,2,3,4\}$ is a subgroup, call it $D$, of the group $S_4$ of all permutations of $\{1,2,3,4\}$.
Observe that the permutations $(1,2,3,4)$ and $(1,3)$ are isometries. (Geometrically, viewing $1,2,3,4$ as vertices of a square, one is a $90^\circ$ rotation, the other is reflection in a diagonal.) In other words, $(1,2,3,4)$ and $(1,3)$ are elements of $D$. Inasmuch as $\langle(1,2,3,4),(1,3)\rangle$ is defined as the smallest subgroup of $S_4$ containing $(1,2,3,4)$ and $(1,3)$, it follows that $\langle(1,2,3,4),(1,3)\rangle\subseteq D.$
To finish the proof, all we have to show is that $D$ is a proper subgroup of $S_4$. For that, it suffices to observe that the permutation $(1,2)$ is not an isometry.
A: Let $g=(1\,2\,3\,4)$, and $h=(1\,3)$. So the order of $g$ is $4$ and the order of $h$ is $2$.
We have $gh=(1\,4)(2\,3)$, so $(gh)^2=e$. Rather, $ghg=h$. That means that any word consisting of $g^i$s ($0<i<4$) and $h^1$s can be reduced to not contain the subword $ghg$. If you have an $h$ surrounded by $g$'s on either side, you can reduce the $ghg$ to just $h$.
So there are not many words left to consider. 


*

*No $h$ at all: $e, g, g^2, g^3,$

*One $h$ at the end of the word: $h, gh, g^2h, g^3h,$

*One $h$ at the beginning of the word: $hg, hg^2, hg^3$

*An $h$ at both beginning an end: $hgh, hg^2h, hg^3h$


Any longer word would necessarily have $ghg$, $h^2$, or $g^4$ as a subword. This is only $14$ words, and that is enough to reach your goal that this subgroup is not the entire $S_4$. (In fact, it only has $8$ elements, since there are duplicate representations here.)

Alternatively, if you have studied the dihedral groups, you might know that $\left\langle g,h\mid g^4=h^2=ghgh=e\right\rangle$ is a construction of the dihedral group of order $8$. So this group has order dividing $8$. Since it clearly has order greater than $4$, it actually has order $8$.
A: Define $A = \langle (1\ 2\ 3\ 4)\rangle = \langle a\rangle$ and $B = \langle (1\ 3)\rangle = \langle b\rangle$.
Note that $A \cap B = \{e\}$.
Note as well that $ba = (1\ 3)(1\ 2\ 3\ 4) = (1\ 2)(3\ 4) = (1\ 4\ 3\ 2)(1\ 3) = a^{-1}b$.
This, in turn, implies that $ba^i = a^{-i}b$, so $BA \subseteq AB$.
Since $|AB| = |BA| = 8$, we conclude $AB$ is a subgroup of $S_4$ of order $8$.
Since $\langle a,b\rangle = \langle A,B\rangle = AB$ (since any subgroup of $S_4$ generated by $A$ and $B$ must contain $AB$ by closure), we conclude $a$ and $b$ do not generate $S_4$.
