Chapter 11, Section D, Problem 5 (from the 2nd edition)
Let $G$ be a group and let $a,b \in G$. Prove the following:
Let ord($a$) = $n$, and suppose $a$ has a $k\text{th}$ root, say $a = b^k$. Then $\langle a \rangle = \langle b \rangle$ iff $k$ and $n$ are relatively prime.
Issue
It's true that $\langle a \rangle = \langle b \rangle \implies k$ and $n$ are relatively prime. However, the converse is false. Here is a counterexample. Let $a = 2$ and $b = 1$ in $\mathbb{Z_6}$ so that $a = b^2$. Then ord($a$) = $3$ = $n$, and $k = 2$. $2$ is relatively prime with $3$, but $\langle 2 \rangle \neq \langle 1 \rangle$.
The provided solution in mstemper says "the same as the previous exercise" which is false since they are two different problems, and problem 4 is simple to prove.
Chapter 11, Section D, Problem 5 (from the 2nd edition)
Let$G$ be a group and let $a,b \in G$ . Prove the following:
Let ord($a$ ) = $n$ , and suppose $a$ has a $k\text{th}$ root, say $a = b^k$ . Then $\langle a \rangle = \langle b \rangle$ iff $k$ and $n$ are relatively prime.
Issue
It's true that$\langle a \rangle = \langle b \rangle \implies k$ and $n$ are relatively prime. However, the converse is false. Here is a counterexample. Let $a = 2$ and $b = 1$ in $\mathbb{Z_6}$ so that $a = b^2$ . Then ord($a$ ) = $3$ = $n$ , and $k = 2$ . $2$ is relatively prime with $3$ , but $\langle 2 \rangle \neq \langle 1 \rangle$ .
The provided solution in mstemper says "the same as the previous exercise" which is false since they are two different problems, and problem 4 is simple to prove.