Sub division rings of division rings Below $^\ast$ denotes "nonzero elements of".
There is a problem in Jacobson's Basic Algebra 1, there is a problem to this effect: if $S$ is a subdivision ring of $\mathbb{H}$ such that $S^\ast$ is a normal subgroup of $\mathbb{H}^\ast$, then $S=\mathbb{H}$ or $S\subseteq\mathbb{R}$.
The solution that came to mind relies a bit on $\mathbb{H}$ having a norm into $\mathbb{R}$, and probably the behavior of $i,j$ and $k$.
I couldn't see a proof for general division rings, though. Does anyone know if it's true for division rings, or have a counterexample?
My gut tells me to lean toward the latter, but I have been bitterly disappointed by my gut feelings before...

Update: Two things have happened: Rankeya has given a valid answer to the written question, but I realize now I was too vague. Secondly, I looked up the correct exercise in Jacobson and found that the following exercise is precisely to show that it does hold for all division rings. Stupid gut feelings...
I'm accepting this answer and reposting the correct question.
 A: There is a celebrated theorem due to Brauer-Cartan-Hua )independently) that asserts that: If D is a division ring with center F, then every subdivision ring H which H* is normal in D* is either central or the whole division ring D.
Your question is an special case of this theorem.
The proof of Cartan-Brauer-Hua theorem by Richard Brauer.
A: Cartan-Brauer-Hua Theorem: Let $D$ be a division ring and $S$ be a division subring of $D$. If $S$ is stabilized by every map $x\mapsto dxd^{-1},d\in D^*$, this is, $\{dxd^{-1}|d\in D^*\}\subset S$ for every $x\in S$, then either $S=D$ or $S\subset C(D)$.
Proof by Loo-Keng Hua: 
Hint: If $ab\ne ba$ then $a=[b^{-1}-(a-1)^{-1}b^{-1}(a-1)][a^{-1}b^{-1}a-(a-1)^{-1}b^{-1}(a-1)]^{-1}$.
Proof of Hint: If $ab\ne ba$ then $a,b\ne 0,1$.
$$b^{-1}a-b^{-1}=b^{-1}(a-1)=(a-1)(a-1)^{-1}b^{-1}(a-1)$$
$$=a(a-1)^{-1}b^{-1}(a-1)-(a-1)^{-1}b^{-1}(a-1).$$
Then $$b^{-1}a-a(a-1)^{-1}b^{-1}(a-1)=b^{-1}-(a-1)^{-1}b^{-1}(a-1),$$
$$a[a^{-1}b^{-1}a-(a-1)^{-1}b^{-1}(a-1)]=b^{-1}-(a-1)^{-1}b^{-1}(a-1).$$
Since $ab\ne ba$, we have $(a-1)b\ne b(a-1)$ and $b^{-1}-(a-1)^{-1}b^{-1}(a-1)\ne 0$. Hence $$a=[b^{-1}-(a-1)^{-1}b^{-1}(a-1)][a^{-1}b^{-1}a-(a-1)^{-1}b^{-1}(a-1)]^{-1}.$$
Proof: If there exists an element $s\in S$ such that $ds\ne sd$, then $d\in S$ by the Hint. Hence for any $s\in S$ and any $d\in D-S$, $sd=ds$.
Suppose that $S$ is not contained in $C(D)$. There exist $s_1,s_2\in S$ such that $s_1s_1\ne s_1s_2$. 
If $S\ne D$, there is an element $a\in D-S$. Since $as_1$ also belongs to $D-S$, we have $(as_1)s_2=s_2(as_1)$ $=(s_2a)s_1=(as_2)s_1$, which contradicts $s_1s_1\ne s_1s_2$. 
Hence $S=D$.
The link to the proof by Loo-Keng Hua.
Proof by Richard Brauer:
If $d\in D, s\in S$, there exist $s_1,s_2\in S$ such that $$ds=s_1d,$$ $$(d+1)s=s_2(d+1).$$
Then $(d+1)s-ds=s_2(d+1)-s_1d$, this is, $$s-s_2=(s_2-s_1)d.$$
If $s_2-s_1\ne 0$ then $d\in S$. Therefore, if $d\not\in S$, then $s_2=s_1$ and $s=s_2$. Thus $s=s_1$, this is, $ds=sd$. Hence for any $s\in S$ and any $d\in D-S$, $sd=ds$.
Suppose that $S$ is not contained in $C(D)$. There exist $s_1,s_2\in S$ such that $s_1s_1\ne s_1s_2$. 
If $S\ne D$, there is an element $a\in D-S$ and $as_2=s_2a$. Since $a+s_1$ also belongs to $D-S$, we have $(a+s_1)s_2=s_2(a+s_1)$, which contradicts $s_1s_1\ne s_1s_2$. 
Hence $S=D$.
The link to the proof by Richard Brauer.
A: Counterexample: Take the function fields $\mathbb{R}(x)$ and $\mathbb{R}(x,y)$. Since they are fields, in particular they are division rings. The former is a subfield of the latter.  That  $\mathbb{R}(x)^*$ is a normal subgroup of $\mathbb{R}(x,y)^*$ follows from commutativity of multiplication. But, $\mathbb{R}(x) \neq \mathbb{R}(x,y)$ and $\mathbb{R}(x)$ is not a subset of $\mathbb{R}$.
