If Every Nonidentity Element of a Group has Order 2, then it’s an Abelian Group
Problem 212
Let $G$ be a group. Suppose that the order of nonidentity element of $G$ is $2$.
Then show that $G$ is an abelian group.
Let $G$ be a group. Suppose that the order of nonidentity element of $G$ is $2$.
Then show that $G$ is an abelian group.
Let $G$ be a finite group of order $n$ and let $m$ be an integer that is relatively prime to $n=|G|$. Show that for any $a\in G$, there exists a unique element $b\in G$ such that
\[b^m=a.\]
Read solution
Let $G$ be a finite group of odd order. Assume that $x \in G$ is not the identity element.
Show that $x$ is not conjugate to $x^{-1}$.
Read solution
Determine whether a group $G$ of the following order is simple or not.
(a) $|G|=100$.
(b) $|G|=200$.
Read solution
Let $p$ be a prime number. Suppose that the order of each element of a finite group $G$ is a power of $p$. Then prove that $G$ is a $p$-group. Namely, the order of $G$ is a power of $p$.
Add to solve later