This seemingly elementary question was asked in Mathematics StackExchange.com: https://math.stackexchange.com/q/4779592/37763. It got upvotes, but no answers or comments, and so I ask it here.

Let $G$ be a simply connected absolutely simple algebraic group of one of the types $^1{\sf A}_{n-1}$ (inner) or $^2{\sf A}_{n-1}$ (outer) over a field $k$. All such groups are described on page 55 of Tits, Classification of algebraic semisimple groups, Proc. Sympos. Pure Math. 9 (Boulder), 1966, pp. 33-61. The descriptions are as follows:

Type $^1{\sf A}_{n-1}$: Special linear group ${\rm SL}_m(D)$, where $D$ is a central division algebra of degree $d$ over $k$, and $n=md$.

Type $^2{\sf A}_{n-1}$: Special unitary group ${\rm SU}_m(D,h)$, where $D$ is a central division algebra of degree $d$ over a quadratic extension $K$ of $k$ with an involution of the second kind $\sigma\colon D\to D$ such that $k=\{x\in K\ |\ x^\sigma=x\}$, $$h\colon D^m\times D^m\to D$$ is a nondegenerate hermitian form relative to $\sigma$, and $n=md$.

Question.I am looking for a down-to-earth proof that all such groups are indeed of the form either ${\rm SL}_m(D)$ or ${\rm SU}_m(D,h)$. (For me, the Book of Involutions is not down-to-earth.)

I know that my group becomes ${\rm SL}_n(\bar k)$ over an algebraic closure $\bar k$ of $k$.