7
$\begingroup$

Let $K$ be an algebraically closed field. Let $SL(n,K), GL(n,K)$ denote the special linear group and general group respectively, and $D(n,K)$ is the diagonal subgroup of $GL(n,K)$.

Then $SL(n,K) \cap D(n,K)$ must be a torus, i.e. a connected diagonalizable subgroup, of $SL(n,K)$. But, must it be maximal?

When char$K=0$, this seems to be the case. When $K=\mathbb{Z}/(2)$ and $n=2$, $SL(2,K) \cap D(2,K)=\{e\}$, here, $e$ denotes the identity element of the two groups. And $e$ is the only diagonalizable matrix in $SL(2,K)$, so this is verified.

But how to prove that this is true in general? And if is not true, what is its counterexample?

Many thanks~

1 Answers 1

5

Let $T$ be a torus in $SL(n,K)$. Then $T$, considered as a subset of $M(n\times n,K)$, is a commuting set of diagonalizable matrices and as such is simultaneously diagonalizable, i.e. there is a $g\in GL(n,K)$ with $gTg^{-1}\subseteq SL(n,K)\cap D(n,K)$. Hence the dimension of any torus is bounded by the dimension of the intersection in question which shows that this intersection is indeed a maximal torus.

  • 0
    @Riccardo Because $SL(n,K)\subset GL(n,K)$ is a normal subgroup.2014-12-14