Prove that $S_4$ is isomorphic to a presentation

Question

I would like to prove that $G=\langle a,b \, | \, a^2,b^4,(ab)^3\rangle \cong S_4$. I tried to list out all the elements in the group presentation and show that it is isomorphic to $S_4$, but it was too tedious. I've also tried to use the homomorphism $\phi:F(a,b)\rightarrow S_4$ where $\phi(a)=(12)$ and $\phi(b)=(1234)$. However I do not know how to show that the normal closure of the relator is actually the kernel of this map. I've also found a different post where this isomorphism is proved by assuming that $S_4$ is isomorphic to another presentation with $3$ generators. But to show that $S_4$ is isomorphic to $\langle x,y,z \mid x^2=y^3=z^4=xyz=1 \rangle$ is also not easy.

Are there any clever ways to prove this isomorphism?

Answer 1

The relation $(ab)^3=e$ implies that $aba = b^{-1}ab^{-1}$, and since $a^2=e$, this is equivalent to $ab^{-1}a = bab$. Written this way we see that $bab$ has order $4$, and its square is equal to $bab^2ab$ and also $ab^2a$, hence $(ab^2a^{-1})b = b^{-1}(ab^2a^{-1})$. Thus we have

$$ \begin{split} i)&\quad ab^3a = bab; \\ ii)& \quad aba = b^3ab^3; \\ iii)& \quad (ab^2a)b = b^3(ab^2a). \end{split} $$

Now an element of $G$ can be written as a word in the generators $a$ and $b$, which because of the orders of $a$ and $b$ must have the form $e; ab^{k_1}ab^{k_2}\cdots$; or $b^{k_1}ab^{k_2}a\cdots$, where the $k_i$s can be read modulo $4$.

But now by considering relations $i)$ to $iii)$ above, we see that we can use $i)$ or $ii)$ to reduce the number of $a$s in the string whenever there are at least two, unless they are separated by $b^2$. In that case, however, we may use $iii)$ to conjugate it past the power of $b$ preceeding or following it, resulting in a cancellation of an $a^2$ unless $(ab^2a)$ contains all instances of $a$ in the word, in which case we may still conjugate it past any power of $b$ preceeding it and thus may assume any word containing $ab^2a$ has the for $ab^2ab^{k_1}$, for $k_1 \in \{0,1,2,3\}$. It follows that every element in $G$ can be written as a word in $a$ and $b$ with $0,1$ or $2$ $a$s as follows: $$ b^{k_1}, b^{k_1}ab^{k_2}, ab^2ab^{k_1}, \quad k_1,k_2 \in \{0,1,2,3\}. $$ and hence $|G|\leq 24$.

Now if we set $\phi(a)=(12)$, $\phi(b)=(1234)$ then $\phi(a)\phi(b) = (12)(1234) = (234)$ has order $3$, hence it satisfies the relations of $G$ and this $\phi$ extends to give a homomorphism from $G$ to $S_4$. Since $\text{im}(\phi)$ contains elements of orders $2,3$ and $4$, it is clear that $12$ divides $|\phi(G)|$ so that $\phi(G)= \text{Alt}_4$ or $S_4$, and since $(12)\in \phi(G)$ it follows that $\phi(G) = S_4$ and $\phi$ is surjective.

Thus since $|G|\leq 24$ and there is a surjective homomorphism from $G$ to $S_4$, it follows that $G \cong S_4$ as required.

Answer 2

To expand on the approach given by Derek Holt in a comment: Let $G$ be the group with given presentation, and we will consider the cosets of $\langle b \rangle$. If as expected $G \simeq S_4$ with $a \mapsto (12), b \mapsto (1234)$, then observe that each coset of $\langle b \rangle$ should have a unique element whose image fixes 4. In particular, if $\sigma(i) = 4$, then $\sigma \cdot (1234)^i$ fixes 4.

Thus, we can expect that the cosets of $\langle b \rangle$ will be: $$\langle b \rangle, e \mapsto () \\ a \cdot \langle b \rangle, a \mapsto (12) \\ (ab) a (ab)^{-1} \cdot \langle b \rangle, (ab) a (ab)^{-1} \mapsto (13) \\ b a b^{-1} \cdot \langle b \rangle, b a b^{-1} \mapsto (23) \\ b^{-1} (ab) b \cdot \langle b \rangle, b^{-1} (ab) b \mapsto (123) \\ b^{-1} (ab)^2 b \cdot \langle b \rangle, b^{-1} (ab)^2 b \mapsto (132). $$

Now, with this setup, what remains to show is that the given relations are enough to show this exhausts all the cosets of $\langle b \rangle$. For this, since $a^{-1} = a$ and $b^{-1} = b^3$, it suffices to show that multiplying any of the given cosets on the left either by $a$ or by $b$ gives another coset in the list. Furthermore, with the expectation that the given map $G \to S_4$ is an isomorphism, we can identify ahead of time which coset we expect the result to be in each case. For example, since $(1234)(23) = (124)$ and $(124)(1234)^2 = (124)(13)(24) = (132)$, we expect that we should have $b [bab^{-1}] = b^{-1}(ab)^2 b \cdot b^2$, from which we will be able to conclude that multiplying the coset $bab^{-1} \cdot \langle b \rangle$ on the left by $b$ will give the coset $b^{-1} (ab)^2 b \cdot \langle b \rangle$.

At this point, admittedly we have 12 cases to check, and each one could end up being ugly to write out. On the other hand, this is a finite amount of work that's left, and each case could conceivably be computer-checked (either using a naive search through the space of expressions giving the normal closure of the relations, or some more advanced algorithm from a CAS).

Once we have established this fact about the cosets, it follows that $[G : \langle b \rangle] \le 6$, and since also $|\langle b \rangle| \le 4$, we get $|G| \le 24$. On the other hand, since $a \mapsto (12), bab^{-1} \mapsto (23), b^2ab^{-2} \mapsto (34)$ generate $S_4$, the morphism is surjective so $|G| \ge 24$. Therefore, $|G| = 24 = |S_4|$; and so the surjective morphism must be an isomorphism.

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In addition to the observation that we have 12 cases to check which might not be easy to do by hand, this does also have the disadvantage that the representatives are not as nice-looking as they are in the solution by krm2233. On the other hand, the setup does show exactly what relations we need to prove without much trial and error being needed, and this approach could potentially be generalized to other similar situations.

Answer 3

This is just an addendum to Daniel Schepler's answer. Although there are $12$ cases to check, most of them are very easy, and none are seriously difficult.

Let $H = \langle b \rangle$. To prove that $|G:H| \le 6$, we will prove that $$G = H \cup Ha \cup Hab \cup Haba \cup Hab^2 \cup Hab^2a$$ and to show that, it is enough to show that, for each of these six cosets $Hg$, $Hga$ and $Hgb$ are equal to one of the six cosets.

These are all straightforward using $a^2=1$ except for the following three.

$(Haba)b = Hb^{-1}a^{-1} = Ha^{-1} = Ha$.

$(Hab^2)b = Hab^{-1} = Hbaba = Haba$.

$(Hab^2a)b = Habbab = Habab^{-1}a =Hb^{-1}ab^{-2}a = Hab^2a$.