To show that $\mathbb{H}$ maps onto $\mathrm{Aut}(\mathbb{H})$, where $\mathbb{H}$ is the Hamilton quaternions, I thought it'd be pertinent to show first that the subgroup of inner automorphisms of $\mathbb{H}$ (which is in fact the entire group of automorphisms, since all the automorphisms end up being inner by Skolem-Noether, though I don't have enough knowledge of algebra to understand anything about this theorem other than its statement) acts transitively on purely imaginary unit quaternions; then to show that the stabilizer of one of these purely imaginary units-- say $i$-- acts transitively on all purely imaginary units orthogonal to it.

Finally, showing that fixing $i$ and some purely imaginary unit orthogonal to it gives only the trivial automorphism shows that all automorphisms are inner, and hence $\mathbb{H}^{*}$ maps onto $\mathrm{Aut}(\mathbb{H})$.

However I can't find an elegant way to solve the first two parts of this-- that is, a proof of this special case of Skolem-Noether with more elementary methods. Any hints would be appreciated.

EDIT: I realize that the purely imaginary quaternions give the unit sphere. Perhaps a proof along these steps using that would be helpful? What exactly would conjugation constitute on the unit sphere? What would an orthogonal unit mean on the unit sphere? Even answering these things would be very helpful.

  • $\begingroup$ So $\operatorname{Aut}(\Bbb H)$ here just means the $\Bbb R$ algebra automorphisms? If so then the Skolem-Noether theorem says exactly that there is an onto homomorphism of $\Bbb H^\ast$ to $\operatorname{Aut}(\Bbb H)$, as you started to say. Are you asking for a proof of the S-N theorem? $\endgroup$
    – rschwieb
    Sep 3, 2014 at 0:39
  • $\begingroup$ @rschwieb No, I'm asking for hints on proving the intermediate steps I mentioned, without the assistance of anything as high up as Skolem-Noether. $\endgroup$
    – user151882
    Sep 3, 2014 at 0:41
  • $\begingroup$ Ah, OK, so a proof of this special case with more elementary methods. That makes sense. If you feel like it, you might add that to the end of the problem statement :) $\endgroup$
    – rschwieb
    Sep 3, 2014 at 0:43

1 Answer 1


The magic formula is

$$Ψ_g(h) = g\,h\,g^{−1}\ ,$$ that evidently provides a group homomorphism from the multiplicative group ℍ× of quaternions to the group Aut(ℍ) of automorphisms of quaternions as a real algebra or, the same, as a topological ring (note that I won’t consider seriously “discontinuous automorphisms” and other set-theoretical perversions). Or, in original poster’s words, all (real☺) automorphisms are inner.

Two more remarks:

  • Ψ has non-trivial kernel ℝ×, the centre of ℍ×.
  • Surjectivity of Ψ follows from Aut(ℍ) ≃ SO(3) isomorphism.

The latter is an exercise that can be performed with vector calculus only; original poster almost got it themselves. Hint: use scalar+vector representation of quaternions and vector product of Euclidean 3-vectors.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .