Let $G$ be a linear algebraic group and $T\subset B\subset G$ where $T$ is torus and $B$ is a Borel subgroup of $G$.
Suppose that $T$ is maximal in $B$. Then, how can I show that $T$ is also maximal in $G$?
Many books take this for granted, and I don't know how i can show this.
Should I use some fact about dimension? or is there some easier way?
Upon further reflection, I realized that it was a rather obvious question.
Let $B$ be a Borel subgroup of $G$. Then we have $\mathrm{Rank}(G)=\mathrm{Rank}(B)$ ($\because$ $\exists$ $T$:maximal torus in $G$. $T\subset B'$ for some Borel subgroup $B'\subset G$. $xB'x^{-1}=B$ for some $x\in G$. Then $xTx^{-1}$ is maximal in $G$ and in $B$.)
Let $T\subset B$ be a maximal torus in $B$ where $B$ is a Borel subgroup of $G$. Then $\dim(T)=\mathrm{Rank}(G)$. There exists a maximal torus $T'$ in $G$ such that $T\subset T'$. Since $\dim(T)=\dim(T')$ and $T, T'$ are connected, we see that $T=T'$.