TITLE: What are relations of $S_4$, if generators are $a=(12)$, $b=(1234)$?
QUESTION [7 upvotes]: I think, relations could be $a^2=e$, $b^4=e$, $ababab=e$, $b^2 a b^2 a b = b a b^2 a$.
By two first relations and $b^3 = ababa$ (it is result of our relations),  we can present any element that is generated by $a$ and $b$, as a sequence, where there are not more than one consecutive $a$ and not more than two consecutive $b$. By the last relation we can put all squares of $b$ to the beginning of sequence. We can remove  3 or more consecutive $ab$ by third relations. And also we have $ab^2ab^2ab^2ab^2=e$ as result of relations, so we can remove 4 or more of consectutive $ab^2$ from the beginning of sequence.
But, I'm not sure that we need exactly 4 relations, that there are not  3 relations that would be sufficient.

REPLY [2 votes]: Here is another proof that $G = \langle x,y | x^2=y^4=(xy)^2=1 \rangle$ presents $S_4$.
Since we know that $S_4$ is a homomorphic image of $G$, we have $|G| \ge 24$, so it is enough to prove that $|G| \le 24$.
Let $H = \langle y \rangle \le G$. It is enough to prove that $|G:H| \le 6$, and to do that we will prove that $G = \bigcup_{Hg \in C} Hg$, where $$C = \{H,Hx,Hxy,Hxy^{-1},Hxy^2,Hxy^2x\}.$$
Since $G$ is generated by $x$ and $y$, it is enough to prove that $Hgx \in C$ and $Hgy \in C$ for all $Hg \in C$.
Most of these are straightforward.
Note that $xyx = y^{-1}x^{-1}y^{-1}=y^{-1}xy^{-1}$ so $Hxyx = Hxy^{-1}$ and similarly $Hxy^{-1}x = Hxy$.
Finally $$xy^2xy = xy(yxy) = xy(xy^{-1}x)= (xyx)y^{-1}x = y^{-1}xy^{-1}y^{-1}x = y^{-1}xy^2x,$$
so $Hxy^2xy = Hxy^2x$.