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Here is an idea I've been working on for self study.

Suppose $S$ is a division subring of $\mathbb{H}$ (the quaternions, viewed as a subring of $M_2(\mathbb{C})$), which is stabilized by the maps $x\mapsto dxd^{-1}$ for all $d\neq 0$ in $\mathbb{H}$. Then either $S=\mathbb{H}$, or $S$ is contained in the center.

I suppose that $S\neq\mathbb{H}$. I define $\varphi_d\colon S\to S$ to be the conjugation map $x\mapsto dxd^{-1}$. Evidently, I find that each $\varphi_d$ is a bijection on $S$, as it is injective, and for any $y\in S$, $d^{-1}yd\in S$ and is the desired preimage for $\varphi_d$.

I also proved that $Z(\mathbb{H})=\mathbb{R}$. I don't know how to show $S\subset Z(\mathbb{H})$ when $S\neq\mathbb{H}$. I want to show $\varphi_d=\mathrm{id}_S$ for all $d\neq 0$ in $\mathbb{H}$, but I don't know how to prove that. Does anyone have any hints or suggestions on what to do?

I assume that $S$ contains at least one imaginary element $ai+bj+ck$. Conjugating by $i$ shows $ai-bj-ck\in S$, so $2ai\in S$. Similarly, $2bj,2ck\in S$. Is it possible to scale these to any real coefficient to conclude that $S$ contains all imaginary elements?

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Show that if $S$ contains one imaginary element it contains all of them. (It would be useful to know that conjugation by the invertible elements of $\mathbb{H}$ acts transitively on the elements squaring to $-1$.) – Qiaochu Yuan Jun 14 '12 at 23:23
To show transitivity, there is a geometric interpretation for conjugation on $S^2 = \{ai+bj+ck \in \mathbb{H} \, | \, a^2+b^2+c^2=1 \}$. Conjugating by a unit quaternion $q = \cos (\alpha/2) + \vec{u} \sin(\alpha/2)$ is a rotation about $\vec{u}$ through an angle $\alpha$, and it follows that the conjugation action is transitive on $S^2$. See… for more information. – Logan Maingi Jun 14 '12 at 23:48
Thanks @QiaochuYuan and Logan, I was at least able to conclude that $2ai,2bj,2ck\in S$. – Noomi Holloway Jun 15 '12 at 0:34
up vote 3 down vote accepted

Take $a\in\mathbb{H}\setminus S$, and take $s\in S^*$. I claim that the commutator $[s,a]=sas^{-1}a^{-1}=1$.

Notice that $a$ is invertible as it is nonzero, and since $a\neq -1$, $a+1\neq 0$, and is thus invertible also. So $[s,a]$ and $[s,a+1]$ exist.

Since $S$ is closed under conjugation, it follows that $[s,a],[s,a+1]\in S^*$. Quick computation shows $$ [s,a+1](a+1)=[s,a]a+1 $$ which implies $$ ([s,a+1]-[s,a])a=1-[s,a+1]. $$ If $[s,a+1]\neq[s,a]$, then $([s,a+1]-[s,a])^{-1}$ exists and is in $S$, so $$ a=([s,a+1]-[s,a])^{-1}(1-[s,a+1])\in S, $$ a contradiction since $a\in\mathbb{H}\setminus S$. So necessarily $[s,a+1]=[s,a]$, which simplifies to imply $[s,a]=1$, or $sa=as$. Thus every element of $S^*$ commutes with every elements of $\mathbb{H}\setminus S$. Not take any $s,t\in S^*$, then if $a\in\mathbb{H}\setminus S$, $a+t\in\mathbb{H}\setminus S$. Then $$ st=s(a+t-a)=s(a+t)-sa=(a+t)s-as=ts $$ where we use the fact that $s$ commutes with $a+t$ and $a$ as found above. So every element of $S^*$ also commutes with every element of $S^*$. Everything still works when considering $0$, so it follows that $S\subset Z(\mathbb{H})$.

Note that nothing inherently special about the quaternions was used here except that $\mathbb{H}$ is a division ring, so this result holds for any division subring of a division ring which is closed under conjugation from the overring.

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Aha, glad I stumbled across this! I had wondered about the answer after reading the question for division rings in Jacobson's book. – rschwieb Dec 3 '12 at 0:32
@rschwieb I hope it helped. I've been reading through Jacobson's Algebra I on and off for about 6 months now, which is why I had this answer prepared. – yunone Dec 3 '12 at 5:07
Actually I've avoided reading this in detail because I still want to work on it... but it's good to know the answer exists somewhere :) – rschwieb Dec 3 '12 at 14:06

After working on it with a friend for a while, we came up with a proof. Upon looking up the answer, we found we had reproduced the ideas of Brauer's proof.

Let $S\subseteq D$ be a division ring extension with $S\neq D$, and suppose that $xSx^{-1}\subseteq S$ for every nonzero $x\in D$. We will show that $S$ is central in $D$.

Lemma 1: If $x\in D\setminus S$ and $s\in S$, then $x$ commutes with $s$.

Proof: Since $S$ is additively closed and $1\in S$, we find that $x+1\notin S$.

Set $t=xsx^{-1}-(x+1)s(x+1)^{-1}\in S$ and calculate that: $$ t(x+1)=xsx^{-1}-s\in S $$

Note though that if $t$ were nonzero, then $t(x+1)\in S$ would imply that $(x+1)\in S$, so it must be that $t=0$. Thus, $xsx^{-1}-s=0$, and after rearrangement, $x$ commutes with $s$.

Lemma 2: If all elements of $S$ commute with all elements outside of $S$, then $S$ is commutative.

Proof: Given $x\in D\setminus S$, and $s,t\in S$:

$$ xst=(xs)t=t(xs)=(tx)s=(xt)s=xts $$ Cancelling $x$ on the left, $st=ts$.

Thus, everything in $S$ commutes with everything in $D$, i.e. $S$ is central.

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