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.


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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Aug 8, 2020 at 7:56
  • $\begingroup$ @HallaSurvivor No it doesn't. $\endgroup$ Aug 8, 2020 at 11:27
  • $\begingroup$ @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)] $\endgroup$ Aug 8, 2020 at 11:44
  • 2
    $\begingroup$ 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! $\endgroup$
    – Derek Holt
    Aug 8, 2020 at 12:18

1 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.)


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