Let $a=b^k$. Prove that $\langle a\rangle=\langle b\rangle$ iff $\gcd(k, \operatorname{ord}(a))=1$.

Question

This is from Pinter's Book of Abstract Algebra Chapter 11, Exercise D5.

Let $n=\operatorname{ord}(a)$.

I think I can prove the $\Rightarrow$ direction: since $a^r$ generates $\langle a \rangle$ iff $\gcd(r, n)=1$, hence $a=b^k$ generates $\langle b \rangle$. Since $a$ generates $\langle a\rangle$ and $\langle b \rangle$, the equality holds.

But I have trouble proving the $\Leftarrow$ direction. What I can see so far:

$\langle a\rangle\subseteq\langle b\rangle$ because $a = b^k\tag 1$

$n\mid\operatorname{ord}(b)\tag 2$ because the order of cyclic subgroup $A$ of cyclic group $B$ divides the order of cyclic group B. $\operatorname{ord}(b)\mid k n\tag 3$ because $b^{kn}=a^k=e$

Because of $(1)$, we know that $\langle a \rangle=\langle b\rangle$ iff $a$ and $b$ have the same order, so it feels like I just need to tighten $(2)$ and $(3)$, but I am stuck. Any help will be appreciated.

Regarding (1): since $\gcd(k, n)=1$, we can find $c$ that satisfies $(a^c)^k=a$, i.e., by solving $ck \equiv 1 (\mod n)$. However, this result can only be used to reduce this question to (1) if $k$-th root of a is unique in $b$ (in which we can imply $a^c = b$). The later I couldn't prove because we don't know $\operatorname{ord}(b)$ — Risan, Jun 05 '20 at 07:57
I put the links in case some other idea occurs to you. Those might be helpful in future. — PinkyWay, Jun 05 '20 at 08:01
Also, if you want "$\mid$" for divides , use \mid and, if you want $\nmid$, use \nmid. — PinkyWay, Jun 05 '20 at 08:12
Does this answer your question? Prove that $\langle a^n \rangle \bigcap \langle a^k \rangle = \langle a^{lcm (n,k)} \rangle$ — Chain Markov, Jun 08 '20 at 15:20
It doesn't. That only helps to prove that $\langle b \rangle \cap \langle a \rangle = \langle a \rangle$. We already know this, i.e., $a \subseteq b$, in the question description above. — Risan, Jun 09 '20 at 16:42

score 2 · Answer 1 · answered Jun 09 '20 at 16:56

2

Looks like it is false, which I overlooked because I was under the assumption that it is true.

Counter example: Consider the cyclic subgroup of $\mathbb{Z}_{10}$. Let's say $a = 2$ and $b = 1$. We know that $a = b^2$, $\operatorname{ord}(a) = 5$, and $\gcd(2, 5) = 1$. However, $\langle 1 \rangle = \mathbb{Z}_{10}$ but $\langle 2 \rangle$ consists of all even numbers of $Z_{10}$. Hence $\langle a \rangle \neq \langle b \rangle$.

answered Jun 09 '20 at 16:56

Risan

113

I recently asked this same question. Can I assume that the question from Pinter's book has a typo since the converse doesn't hold? I.e. it is not the case that $\langle a\rangle = \langle b\rangle$ whenever $a = b^k$ and the order of $a$ and $k$ are relatively prime? – Display name Oct 23 '23 at 22:41

Let $a=b^k$. Prove that $\langle a\rangle=\langle b\rangle$ iff $\gcd(k, \operatorname{ord}(a))=1$.

1 Answers1