Hw6 Problem 1

Return to Homework 6, Glossary, Theorems

### Problem 1

Let $G$ be a group and $a\in G$.

(a) If $a^{12} = e$, what can we say about $|a|$?
(b) If $a^m = e$, what can we say about $|a|$?
(c) Suppose that $|G| = 24$ and that $G$ is cyclic. If $a^8 \neq e$ and $a^{12} \neq e$, show that $\langle a \rangle = G$.

Solution

(a) By Theorem 6.14 $|a|$ divides $12$, so $|a| = 1, 2, 3, 4, 6,$ or $12$.

(b) The order of $a$ is a divisor of $m$.

(c) Suppose $|G|=24$ and $G$ is cyclic, then by Theorem 6.14 we know that the order of $a$ divides $|G|$, so $|a| = 1, 2, 3, 4, 6, 8, 12,$ or $24$. However, since $a^8 \neq e$ and $a^{12} \neq e$ $|a| \neq 8, 12$ or any of their divisors. So $|a| \neq 1, 2, 3, 4, 6, 8,$ or $12$. Therefore, $|a| =24$.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License