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.

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    $\begingroup$ In fact $\langle x,y \mid a^2=b^4=(ab)^3=1 \rangle$ is a presentation of $S_4$, so you only need the first three of your relations. $\endgroup$
    – Derek Holt
    Sep 23, 2021 at 21:47
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    $\begingroup$ Your fourth relation is derived from the first three. Since $(ab)^3=1$ and $a^{-1}=a$ it follows that $bab=ab^{-1}a$. Therefore, $$ b^2ab^2ab=b(bab)^2=b(ab^{-1}a)^2=bab^{-2}a=bab^2a. $$ $\endgroup$
    – kabenyuk
    Sep 24, 2021 at 5:20

2 Answers 2


Here is another proof that $G = \langle x,y | x^2=y^4=(xy)^3=1 \rangle$ 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 $Hxy^2xy = Hxy^2x$.

  • $\begingroup$ Yes, thanks, I've corrected it. $\endgroup$
    – Derek Holt
    Jan 6 at 23:01

As Derek Holt mentions in the comments you have $S_4 \cong \langle x,y | x^2 = y^4 = (xy)^3 = 1 \rangle$, so for the generators $a = (1\ 2)$ and $b = (1\ 2\ 3\ 4)$ three relations $a^2 = b^2 = (ab)^3 = 1$ suffice.

Let $G = \langle x,y | x^2 = y^4 = (xy)^3 = 1 \rangle$. In $S_4$ you have a normal subgroup $V \cong C_2 \times C_2$ of order $4$, generated by $b^2 = (1\ 3)(2\ 4)$ and $ab^2a^{-1} = (1\ 4)(2\ 3)$.

So consider in $G$ the subgroup $N = \langle z, w \rangle$ where $z = y^2$ and $w = xy^2x^{-1}$. We would expect that $N$ is a normal subgroup and $N \cong C_2 \times C_2$.

For normality, it is clear that $xzx^{-1} = w$, $xwx^{-1} = z$, and $yzy^{-1} = z$. It remains to check that $ywy^{-1} \in N$. For this, using $(yx)^3 = 1$ and $x^2 = 1$ we find that $yxy = xy^{-1}x$ (as suggested in a comment). Then \begin{align*} ywy^{-1} &= yxy^2xy^3 \\ &= (yxy)(yxy)y^2 \\ &= (xy^{-1}x)(xy^{-1}x)y^2 \\ &= xy^2xy^2 \\ &= wz. \end{align*}

Certainly $N$ has $C_2 \times C_2$ as a quotient. Check that $(zw)^2 = 1$ so in fact $N \cong C_2 \times C_2$: \begin{align*} (zw)^2 &= y^2xy^2xy^2xy^2x \\ &= y(yxy)(yxy)(yxy)yx \\ &= y(xy^{-1}x)(xy^{-1}x)(xy^{-1}x)yx \\ &= y(xy^{-3}x)yx \\ &= yxyxyx = (yx)^3 = 1.\end{align*}

Then $G/N$ is generated by $\overline{x}$, $\overline{y}$ with $(\overline{x})^2 = (\overline{y})^2 = (\overline{x}\overline{y})^3 = 1$, so $G/N$ is a quotient of $S_3$.

Conclude $G \cong S_4$.

  • $\begingroup$ But you also need to prove that $N$ is normal in $G$, which I think reduces to showing that $ywy^{-1} \in N$. $\endgroup$
    – Derek Holt
    Sep 24, 2021 at 8:15
  • $\begingroup$ @DerekHolt: That's right, I will add this detail. Here $V$ being normal implies that $NK$ is normal, where $K$ is the kernel of the map $G \rightarrow S_4$ with $x \mapsto a$ and $y \mapsto b$. But this would not immediately/obviously imply that $N$ is normal (as my answer might seem to suggest). $\endgroup$
    – spin
    Sep 24, 2021 at 12:44
  • $\begingroup$ Anyway, this answer is a little bit of a trick because it relies on knowing the solution already, where the idea of looking for this normal subgroup comes from. Coset enumeration provides a more general method. $\endgroup$
    – spin
    Sep 24, 2021 at 13:03
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    $\begingroup$ It might be a trick, but my standard approach to unseen problems of this type is first to try a completely naive computer calculation, if that doesn't work then think about it and try a less naive computer calculation, etc. If this leads to an answer then look at the answer and see if there is a clever way to do it by hand. If so then of course you just present the final solution and it looks like magic. $\endgroup$
    – Derek Holt
    Sep 24, 2021 at 13:40

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