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Problem 91 (The Order Of An Element Divides The Order Of A Group)

Let $G$ be a finite group. Use Lagrange's theorem to prove the following two corollaries.

  1. Let $a\in G$. The order of $a$ divides the order of $G$.
  2. If $|G|=p$ for some prime $p$, then $G$ is cyclic.

Solution

1. We know that $|a| = |\langle a\rangle|$ from previous problems. Since $\langle a\rangle$ is a subgroup of $G$, we know by Lagrange's Theorem that $|\langle a\rangle|$ divides $|G|$, so $|a|$ divides $|G|$.

2. Let $a\in G$ with $a\neq e$ (note $1$ is not prime, so $G$ must have at least one element not equal to $e$). We know by Lagrange's Theorem that $|\langle a\rangle|$ divides $|G|$. Since $|G|$ is a prime number $p$, we know that the only divisors of $p$ are $1$ and $p$. Since we chose $a$ to not be equal to $e$, we know that $|a| \neq 1$. Then $|\langle a\rangle| = p$ and $\langle a \rangle = G,$ so $G$ is cyclic.

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