# What are relations of $S_4$, if generators are $a=(12)$, $b=(1234)$?

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.

• 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. Sep 23, 2021 at 21:47
• 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, $$b^2ab^2ab=b(bab)^2=b(ab^{-1}a)^2=bab^{-2}a=bab^2a.$$ Sep 24, 2021 at 5:20

## 2 Answers

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 $$Hxy^2xy = Hxy^2x$$.

• Yes, thanks, I've corrected it. 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 $$G \cong S_4$$.

• But you also need to prove that $N$ is normal in $G$, which I think reduces to showing that $ywy^{-1} \in N$. Sep 24, 2021 at 8:15
• @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).
– spin
Sep 24, 2021 at 12:44
• 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.
– spin
Sep 24, 2021 at 13:03
• 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. Sep 24, 2021 at 13:40