Prove that $K$ is a Sylow $p$-subgroup of $T$.

Question

Let $G$ be a finite group and let $T$ be a subgroup of $G.$ Suppose $K$ is a Sylow $p$-subgroup of $G$ such that $K \subseteq T.$

Since $K$ is a Sylow $p$-subgroup of $G$, then $|K|=p^m.$ I want to use LaGrange's theorem in here but I don't know how to connect the dots. Also I know I want to find what the largest power of $p$ that divides $∣T∣$ is but I am confused on how to write this out.

Answer 1

Since the order of $T$ divides the order of $G$ by Lagrange's theorem, the largest power of $p$ that divides the order of $T$ can be no larger than $p^m$ - otherwise we would get that a larger power of $p$ than $p^m$ divides the order of $G$.

From $K \subseteq T$ we get that $|K| = p^m$ divides the order of $T$, thus $p^m$ is the largest power of $p$ dividing the order of $T$, and so $K$ is a Sylow-$p$-subgroup of $T$.

Answer 2

If $r|s$ and $s=p^nt$, with $(p,t)=1$, then if $r=p^nk$, we have $(p,k)=1$, by transitivity of divisibility.

Lagrange's theorem is used to get $r=|T|||G|=s$.