Usage go the Fundamental Theorem of Cyclic Groups

Question

I have $$\langle a^{3}\rangle=\langle b^{10} \rangle \ \text{and} \ |\langle a^{3}\rangle|=12 $$ Then I know that $$|\langle b^{10}\rangle|=12$$ I am trying to use the FTCG that way: $$\langle b^{10}\rangle=\langle b^{n/k}\rangle=\langle b^{n/12}\rangle \implies n=120 \text{ where $n$ is the order of $\langle b\rangle$}$$

Is this a legit use of the theorem (for a finite cyclic group of order n, every subgroup's order is a divisor of n, and there is exactly one subgroup for each divisor) and is the order the subgroup found correctly?

Answer 1

It can't be a legitimate use of the theorem because the order of $b$ could be $24$. For example, in the cyclic group $\langle g \rangle$ of order $72$, take $a = g^2$ and $b = g^3$.

Edit

You are correct that $\langle b^{n/12}\rangle$ is the unique subgroup of $\langle b \rangle$ of order $12$, so $\langle b^{10}\rangle = \langle b^{n/12}\rangle$, but it doesn't follow that $10 = n/12$, only that $\gcd(n,10) = n/12$.

I'm not sure what the best way of solving this is. One way is to let $d = \gcd(n,10)$. Then $n = 12d$, so $$d = \gcd(12d,10) = 2\gcd(6d,5) = 2\gcd(d,5).$$ Now $\gcd(d,5) \in \{1, 5\}$ so $d \in \{2,10\}$ and $n \in \{24,120\}$ and somehow convince yourself that both these values are possible.

It might seem easier just to say $n = 12\gcd(n,10)$ and $\gcd(n,10) \in \{1,2,5,10\}$ so $n \in \{12,24,60,120\}$, but by doing it this way we have obtained some extra possible values of $n$. You can substitute them into $n = 12\gcd(n,10)$ to see that they are not really solutions.