A finite division ring $D$ is a field

Question

7.8.12 Wedderburn: A finite division ring $D$ is a field.

I have understood several theorems from this book (A first course in abstract algebra by Hiram, paley) by my own, but this one is beyond everything. Every time I give a try to understand this proof, I almost lost the track. What I need the most is the understanding the flow of logic which used in the solution. I guess at least that will help me to understand as well as remember the flow when I will give a try to write it. I wish I could write all those lemmas but section 7.8 was here where most of the statement mentioned.

Proof: Clearly, if $D$ has two elements, $D$ is isomorphic to $Z_2$ and therefore it is a field.

If the theorem is not true for all finite division rings $D$, let $D$ be chosen so that $D$ has minimal order among the noncommutative division rings. Thus, every division ring with fewer elements than $D$ is commutative. We shall show that this assumption about $D$ leads to a contradiction.

Since $D$ is not a field, by Lemma 7.8.8. there exists an element $a \in D$ and an integer $r$ satisfying the conclusions of that Lemma. Since $a \notin C$, by Lemmas 7.8.5 and 7.8.6, there exist $x \in D$ and an integer $i$ such that $x a x^{-1}=a^i \neq a$, and clearly $x \notin C$. Since $r$ is a prime, by Exercise 5, Section 4.5, and by Corollary 2.14.2, $x^{r-1} a x^{-(r-1)}=a^{i^{r-1}}=a^{1+ru}=a \cdot a^{r u}$ $=\lambda a$, since $a^{r u}=\lambda$, for some $\lambda \in C$.

By our choice of $a$ and $r, x^{r-1} \notin C$. By Lemma 7.8.7, with $x=b$ and $r-1=s$, we see that $x^{r-1} a \neq a x^{r-1}$. Thus, $x^{r-1} a x^{-(r-1)}=\lambda a$, with $\lambda \in C, \lambda \neq 1$.

Now let $b=x^{r-1}$. Note that $b a \neq a b$. Then $b a b^{-1}=\lambda a$, and since $a^r \in C$, $$ a^r=b a^r b^{-1}=\left(b a b^{-1}\right)^r=(\lambda a)^r=\lambda^r a^r, \text { whence } \lambda^r=1 . $$

I didn't see from where $a^r=b a^r b^{-1}=\left(b a b^{-1}\right)^r$ because here we only have $bab^{-1}=\lambda a$

Thus, $b^r=\lambda^{r} b^r=\lambda^{r}\left(a^{-1} b a \lambda^{-1}\right)^r=\left(a^{-1} b a\right)^r=a^{-1} b^{r} a$, whence $b^r a=a b^r$. By Lemma 7.8.7, $b^r \in C$.

Since $C^*=C-\{0\}$ is cyclic (Theorem 7.7.3), say with generator $\gamma$, and since $a^r$ and $b^r$ are in $C$, there exist integers $n$ and $m$ such that $a^r=\gamma^n$, $b^r=\gamma^m$. Moreover, $r \nmid n$ and $r \nmid m$. For, suppose $n=k r$. Then $a^r=\gamma^n$ $=\gamma^{k r}$, whence $a^r\left(\gamma^{-k}\right)^r=1$ and, since $\gamma \epsilon C,\left(a \gamma^{-k}\right)^r=1$. By Lemma 7.8.9, $a \gamma^{-k}=\lambda^i$, whence $a=\lambda^i \gamma^k \in C$, a contradiction.

Now let $a_1=a^m$ and $b_1=b^n$. Then $a_1^r=a^{m r}=\gamma^{m n}=b^{n r}=b_1^r$. Also, $a_1 b_1 a_1^{-1} b_1^{-1}=a^m b^n a^{-m} b^{-n}$ $$ \begin{aligned} & =a^m\left[b^n a^{-m} b^{-n}\right]=a^m\left(\lambda^{-m n} a^{-m}\right)\\ & =\lambda^{-m n}=\mu \in C \end{aligned} $$ By Lemma 7.8.10. Thus, $$a_1 b_1=\mu b_1 a_1,\text{ with }\mu \neq 1\qquad (*)$$. To see $\mu \neq 1$, suppose $\mu=1$. Then $\lambda^{-m n}=1$, whence $r \mid m n$, which is a contradiction, since $r \nmid m$ and $r \nmid n$, and since $r$ is prime. In addition, $\mu^r=\left(\lambda^{-m n}\right)^r=\left(\lambda^r\right)^{-m n}=1$. Next, using $(*)$, we get

$$ \begin{aligned} \left(b_1^{-1} a_1\right)^r & =\mu^{-(1+2+\cdots+(r-1))} b_1{ }^{-r} a_1{}^{r} \\ & =\mu^{-r(r-1) / 2}, \quad \text { since } a_1^r=b_1^r . \end{aligned} $$

How they get $\left(b_1^{-1} a_1\right)^r =\mu^{-(1+2+\cdots+(r-1))} b_1{ }^{-r} a_1{}^{r}$?

We are about ready to get a contradiction that will establish the theorem.

Case I: $\quad r$ is odd. Then $\mu^{-r(r-1) / 2}=1$, since $\mu^r=1$, and so $\left(b_1^{-1} a_1\right)^r=1$. But by Lemma 7.8.9, $b_1^{-1} a_1=\lambda^i$, for some $i$, whence $a_1=\lambda^i b_1$, but then $a_1 b_1=b_1 a_1$, a contradiction, and the theorem is proved if $r>2$.

How did they conclude $\mu^{-r(r-1) / 2}=1$?

Case II: $\quad r=2$. Since $\mu^2=1, \mu \neq 1$, necessarily $\mu=-1$. But then $a_1 b_1=-b_1 a_1 \neq b_1 a_1$, by $\left({ }^*\right)$. Thus, char $D \neq 2$. By Corollary 7.8.2, with $\alpha=a_1{ }^2=b_1{ }^2 \in C$, there exist $\xi, \eta \in C$ such that $1+\xi^2-\alpha \eta^2=0$. By Lemma 7.8.11, $\left(a_1+\xi b_1+\eta a_1 b_1\right)=0$. But then $$ 0=a_1\left(a_1+\xi b_1+\eta a_1 b_1\right)+\left(a_1+\xi b_1+\eta a_1 b_1\right) a_1=2 a_1^2=2 \alpha \neq 0, $$ since char $D \neq 2$. This contradiction establishes the theorem in case $r=2$, and the proof of Wedderburn's theorem is concluded.

I didn't understand how the contradiction arise; isn't we want to show our $D$ is commutative? Then what the cases $I,II$ doing here?

Most the questions are resolved now, except the highlighted $2$.

Answer 1

Regarding your first question, in Lemma 7.8.8 $C$ is defined to be the center of $D$ and, by the same lemma, you have $a^r\in C$. Then $a^r=a^r1_D=a^rbb^{-1}=ba^rb^{-1}$ where $a^r$ and $b$ commute since $a^r$ is in the center of $D$.
Then you also have $(bab^{-1})^r=(bab^{-1})\dots(bab^{-1})=ba^rb^{-1}$.

Regarding your third question, you have $\mu^{-r(r-1)/2}=(\mu^{r(r-1)/2})^{-1}=1$ since $\mu^{r}=1$ as stated a little above (just under the expression with $(\ast)$).

Second question: induction on $r$. Note that, since $\mu\in C$, by $(\ast)$ you have $a_1b_1^{-1}=\mu^{-1}b_1^{-1}a_{1}$.

$r=1$: obvious.

$r=2$ (not necessary, just to understand how things work): \begin{equation} (b_1^{-1}a_1)^{2}=b_1^{-1}a_1b_1^{-1}a_1=b_1^{-1}\mu^{-1}a_1=\mu^{-1}b_1^{-1}a_1 \end{equation}

$r-1\implies r$: let $k:=-(1+2+\dots (r-2))$. Then:
\begin{align} (b_1^{-1}a_1)^r&=(b_1^{-1}a_1)(b_1^{-1}a_1)^{r-1}=(b_1^{-1}a_1)\mu^{k}b_1^{-(r-1)}a_1^{r-1}=\\ &=\mu^{k}b_1^{-1}a_1b_1^{-(r-1)}a_1^{r-1}=\mu^{k}b_1^{-1}a_1b_1^{-1}b_1b_1^{-(r-1)}a_1^{r-1}=\\ &=\mu^{k}b_1^{-1}\mu^{-1}b_1^{-1}a_1b_1b_1^{-r+1}a_1^{r-1}=\mu^{k-1}b_1^{-1}b_1^{-1}a_1b_1^{-r+2}a_1^{r-1}=\\ &=\mu^{k-1}b_1^{-2}a_1b_1^{-r+2}a_1^{r-1}=\dots=\mu^{k-(r-2)}b_1^{-(r-1)}a_1b_1^{-1}a_1^{r-1}=\\ &=\mu^{k-(r-2)}b_1^{-(r-1)}\mu^{-1}b_1^{-1}a_{1}a_1^{r-1}=\\ &=\mu^{k-(r-1)}b_1^{-(r-1)}b_1^{-1}a_{1}a_1^{r-1}=\mu^{k-(r-1)}b_1^{-r}a_1^{r} \end{align}

I know it looks confused but it was the best I could come up with.