## 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 $x^{-1}$.

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Determine whether a group $G$ of the following order is simple or not.

(a) $|G|=100$.

(b) $|G|=200$.

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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 $p$.

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