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

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.

• Since you have proved that ${\rm Aut}(Q_8)|=24$, you don't necessarily have to prove that $f$ and $g$ generate ${\rm Aut}(Q_8)$. You know that there is a surjective homomorphism from $S_4$ to the subgroup they generate, so you just need to prove that it has trivial kernel. Aug 8, 2020 at 7:56
• @HallaSurvivor No it doesn't. Aug 8, 2020 at 11:27
• @DerekHolt Thanks, I didn't get why such homomorphism must exist though. Assume that the homomorphism is F then without loss of generality assuming F(x) = f and F(y) = g where x = (1 2 3 4) and y = (1 2) , we get Ker F as trivial. [We can choose x because of symmetry between all positions and once x is chosen the only order 2 elements which generate S4 (along with x) are (1 2),(2 3),(3 4),(4 1) hence WLOG we can choose y=(12)] Aug 8, 2020 at 11:44
• The homomorphism exists because the image $x$ and $y$ satisfy the relations of the presentation of $S_4$; i.e. $f^4=g^2=(fg)^3=1$ - you have already remarked on that! The existence of the homomorphism under those conditions is about the single most basic property of groups defined by presentations! Aug 8, 2020 at 12:18

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.
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.)