Prove $\text{Aut}(Q_8)$ is isomorphic to $S_4$ using presentation of groups

Question

We are given that presentation of $S_4$ is $ \langle x,y|x^4=y^2=(x*y)^3=1 \rangle$ and $|S_4| = 24$.

My approach:

I have proven that $|\text{Aut}(Q_8)|=24$ and there exists $f$ and $g$ in $\text{Aut}(Q_8)$ such that $f^4 = \text{Id}$ and $g^2= \text{Id}$ and $(f*g)^3= \text{Id}$ (Where $f(i)=i$,$f(j)=k$ and $g(i)=j,g(j)=i$). If I prove that $f$ and $g$ generate the whole of $\text{Aut}(Q_8)$ then I am done, but this seems rather a tough task. I have speculations that after finding $f$ and $g$ is enough to prove it but I am not sure why.

It would be great if anyone could help.

Edit:

There is a similar question here Automorphism group of the quaternion group

But it doesn't prove by presentation of groups and I want the question to be solved in the way I have approached.

Answer 1

If we continue with the more mechanical approach to this problem, to show that $f$ and $g$ generate $Aut(Q_8)$, it suffices to show that the size of the subgroup that they generate is $24$. Lagrange's theorem implies that the order of an element must divide the order of the group.

Begin by proving the following things:

1. $f$ actually has order 4 (i.e., $f, f^2 \neq Id$).
2. $|\langle f,g\rangle| \geq 12$, i.e., $|\langle f,g\rangle| \in \{12,24\}$.

Now we just need to show that $|\langle f,g\rangle|$ is divisible by 8; you could show this by writing down a subgroup of $\langle f,g\rangle$ that has order 8. (Assuming the conclusion you are trying to prove, you are guaranteed such a subgroup exists by Sylow's first theorem, so this method should be fruitful.)