Every Group of Order 24 Has a Normal Subgroup of Order 4 or 8
Problem 568
Prove that every group of order $24$ has a normal subgroup of order $4$ or $8$.
Add to solve laterProve that every group of order $24$ has a normal subgroup of order $4$ or $8$.
Add to solve laterLet $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer.
Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$.
Then prove that $H$ is a normal subgroup of $G$.
(Michigan State University, Abstract Algebra Qualifying Exam)
Add to solve laterLet $G$ be a finite group of order $231=3\cdot 7 \cdot 11$.
Prove that every Sylow $11$-subgroup of $G$ is contained in the center $Z(G)$.
Let $G$ be a group and let $H$ be a subgroup of finite index. Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.
Add to solve laterLet $G$ be a simple group and let $X$ be a finite set.
Suppose $G$ acts nontrivially on $X$. That is, there exist $g\in G$ and $x \in X$ such that $g\cdot x \neq x$.
Then show that $G$ is a finite group and the order of $G$ divides $|X|!$.