Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Browsing this question: Why are the solutions of polynomial equations so unconstrained over the quaternions?, the pdf linked in the comments says that the infinitely many conjugates of $i$ in $\mathbb{H}$ are roots to $x^2+1$.

I get that they're roots, but how do we know that the conjugacy class of $i$ is in fact infinite?

share|cite|improve this question
Have you tried writing down the conjugacy class of $i$? – Qiaochu Yuan Jul 11 '12 at 17:00
Calculate $$(\cos\alpha +\sin\alpha j)^{-1}i(\cos\alpha +\sin\alpha j).$$ – Jyrki Lahtonen Jul 11 '12 at 17:01

This should be a comment but it came up too long.

I think there's a confusion here: the quaternion $\,i\,$ has one unique quaternionian conjugate in $\,\Bbb H\,$, as we can deduce from this definition

The sense in which the term "conjugate" seems to be used in the OP is that the minimal polynomial of $\,i\,$ over $\,\Bbb H\,$ has infinite other roots, all of which are "conjugate" to (i.e., roots of the same minimal polynomial of) $\,i\,$

So the question doesn't seem to be connected to conjugacy classes a la group theory but with roots of minimal polynomials.

share|cite|improve this answer
Hmm. Are you thinking about conjugates in a group of 8 elements? I was thinking about conjugates in the infinite group of invertible quaternions. Edit: Oh, you were referring to the antiautomorphism of $\mathbb{H}$. Sorry. Still, in the group theoretic sense $i$ has infinitely many conjugates. – Jyrki Lahtonen Jul 14 '12 at 19:23
As far as I know $\,\Bbb H\,$ usually denotes the infinite division ring of quaternions, and $\,Q_8\,$ the group of quaternions of order $\,8\,$...and no: I wasn't referring to group theoretic meaning but to the quaternionian meaning as in the linked definition. – DonAntonio Jul 15 '12 at 0:03
Quite. I realized what you had meant, but decided to leave the comment anyway. Talking about a conjugacy class does suggest that the OP is interested in the group theoretic version of "conjugate". – Jyrki Lahtonen Jul 15 '12 at 6:02

$$(a+bi+cj+dk)^2+1=0$$ $$a^2+1 + b^2i^2+c^2j^2+d^2k^2+2abi+2acj+2adk+(bcij+cbji)+(cdjk+dckj)+(bdik+dbki)=0$$ The last three brackets are all $0$. The $i,j,k$ components must all be zero as well, which we can do by setting $a=0$. We're left with $$1-b^2-c^2-d^2=0$$ which can be satisfied for infinitely many $b,c,d$.

DonAntonio seems to have covered the issue about quaternion conjugates quite nicely, if that's more what you were asking.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.