Proving there is a unique subgroup $H\le S_4$ isomorphic to $G=\langle (1~~3),(2~~3~~4)\rangle\le S_7.$

Question

Suppose $G$ is a subgroup of $S_7$ generated by the elements $a=(1~~3)$ and $b=(2~~3~~4).$ Prove there exists a unique subgroup $H$ of $S_4$ isomorphic to $G.$

My attempt

Both $a$ and $b$ fix the set $\{5,6,7\}$ and hence does every element of $G=\langle a,b\rangle.$

Note that $|a|=2$ and $|b|=3.$

Let $f:G\to S_4,f:(5~~6~~7)\sigma\mapsto\sigma.$ Then, $f$ is a monomorphism and I want to show it is surjective. Since $\operatorname{Im}f\le S_4,$ it would be sufficient to prove $|\operatorname{Im}f|=|S_4|=24.$

If we also denote by $a,b$ the corresponding images (now in $S_4$) of $a,b$ under $f,$ then $$\operatorname{Im}f=\langle\operatorname{id},b,b^2,a,ab,ab^2,ba,bab,bab^2,b^2a,b^2ab,b^2ab^2\rangle.$$ According to this answer, $S_4$ is generated by $4$ - cycles and there are $\binom443!=6$ of those. Also, since conjugated permutations have the same cycle type, I considered just a few elements of $\operatorname{Im}f:$ $$\begin{aligned}ab&=(1~~3)(2~~3~~4)=(2~~1~~3~~4)=(1~~3~~4~~2)\\a(ab)a^{-1}&=a(ab)a=ba=(1~~4~~2~~3)\\b^{-1}(ab)b&=b^2(ab)b=b^2ab^2=(1~~2~~3~~4)\\(ab)^{-1}&=b^{-1}a^{-1}=b^2a=(1~~2~~4~~3)\\(ba)^{-1}&=a^{-1}b^{-1}=ab^2=(1~~3~~2~~4)\\(b^2ab^2)^{-1}&=b^{-2}ab^{-2}=bab=(1~~4~~3~~2).\end{aligned}$$ Those are $6$ distinct $4$ - cycles so $\operatorname{Im}f=S_4.$ I also reached the same conclusion when writing down all the non - repeating elements that I could, more or less, instantly think of: $$\begin{aligned}\text{id}\\b&=(2~~ 3~~ 4)\\b^2&=(2~~ 3~~ 4)(2~~ 3~~ 4)=(2~~ 4~~ 3)\\a&=(1~~ 3)\\ab&=(1~~ 3)(2~~ 3~~ 4)=(2~~ 1~~ 3~~ 4)=(1~~ 3~~ 4~~ 2)\\ab^2&=(1~~ 3)(2~~ 4~~ 3)=(2~~ 4~~ 1~~ 3)=(1~~ 3~~ 2~~ 4)\\ba&=(2~~ 3~~ 4)(1~~ 3)=(1~~ 4~~ 2~~ 3)\\bab&=(2~~ 3~~ 4)(1~~ 3~~ 4~~ 2)=(1~~ 4~~ 3~~ 2)\\bab^2&=(2~~ 3~~ 4)(1~~ 3~~ 2~~ 4)=(1~~ 4)\\b^2a&=(2~~ 3~~ 4)(1~~ 4~~ 2~~ 3)=(1~~ 2~~ 4~~ 3)\\b^2ab&=(2~~ 3~~ 4)(1~~ 4~~ 3~~ 2)=(1~~ 2)\\b^2ab^2&=(2~~ 3~~ 4)(1~~ 4)=(1~~ 2~~ 3~~ 4)\\(ab)^2&=abab=(1~~ 3)(1~~ 4~~ 3~~ 2)=(1~~ 4)(2~~ 3)\\(ba)^2&=baba=(1~~ 4~~ 3~~ 2)(1~~ 3)=(1~~ 2)(3~~ 4)\\(bab)^2&=bab^2ab=(1~~ 4)(1~~ 3~~ 4~~ 2)=(1~~ 3)(2~~4).\end{aligned}$$ There are $15$ of them and $12<15<24.$ However, by Lagrange, they have to generate a subgroup of $24$ elements and there is only one such subgroup in $S_4.$

Is there any other way to solve this problem? This task was on an exam and among easier ones, but I still spent too much time on it so I believe there must be something more obvious that I haven't noticed.

Answer 1

As pointed out by Amateur_Algebraist in the comments, $\langle a, ab\rangle=\langle (1~~3),(1~~3~~4~~2)\rangle=\langle(1~~2),(1~~2~~3~~4)\rangle=S_4,$ which yields the answer.

Answer 2

Clearly, there is some subgroup $H$ of $S_4$ isomorphic to $G$. So, only uniqueness is to show. Assuming uniqueness first, we see that by permuting numbers $1,2,3,4$ that it must be $H = S_4$, since transpositions $(i \ j)$ generate the symmetric groups. So, the only thing to show is, that all transpositions are in $H$. $(1\ 4) = bab^{-1}, (1 \ 2) = b^{-1}ab$. Then $(1 \ 2)b(1 \ 2) = (1 \ 3 \ 4)$, therefore $(3 \ 4) \in H$, and then $(2 \ 3) = b(3 \ 4) \in H$. Finally, $(2 \ 4) = b(2 \ 3) \in H$.