Find all elements $x \in \mathbb{Z}_{96}$ such that $\left<x\right>=\left<12\right>$.
I know the answer of this question, the answer is $\left<12,36,60,84\right>$.
However, I am confused that why the answer is not $\left<12,24,36,48,60,72,84\right>$. Can anyone help me?
On
For $⟨x⟩$ to be equal to $⟨12⟩$, $12$ needs to be an element of $⟨x⟩$ and $x$ needs to be an element of $⟨12⟩$. Clearly $24, 48$ and $72$ are elements of $⟨12⟩$ but $12$ is not an element of $⟨x⟩$ in any of these cases.
For $12$ to be an element of $⟨x⟩$, we need $n \in \mathbb{Z}_{96}$ such that $nx \equiv12 \ (\text{mod} \ 96).$
Take $24$ as an example. We find the congruence $24n \equiv 12 \ (\text{mod} \ 96) $ is not solvable as $\gcd(24, 96) = 24$ and $24$ doesn't divide $12$, therefore $12 \notin ⟨24⟩$. This is the same for $48$ and $72$. In the other cases, the congruence is solvable, so $12 \in ⟨x⟩$.
On
The subgroup $\langle12\rangle$ is generated by $12a$ if and only if $a$ is coprime to $\frac{96}{12}=8$, i.e. odd. Hence, $12=12 \cdot 1$, $36=12 \cdot 3$, $60=12 \cdot 5$, and $84=12 \cdot 7$ are generators, but $24=12 \cdot 2$, $48=12 \cdot 4$, and $72=12 \cdot 6$ are not.
On
$\gcd(12,96) = \gcd(36,96)= \gcd(60,96)=\gcd(84)=12$.
$\gcd(24,96) = \gcd(72,96)=24 \ne 12$
And $\gcd(48,96) = 48\ne 12$.
Can you see why that would make a difference?
(Note every element of $<a>\subset \mathbb Z_{m}$ will be a multiple of $\gcd(a,m)$. So $<24>$ will only contain multiples of $24$ and will not contain any multiples of $12$ that are not multiples of $24$.)
Note that $\left<24\right> = \{0, 24, 48, 72\}$ since $72+24 = 96 \equiv 0 \pmod{96}$, but $\left<12\right> = \{0, 12, 24, 36, 48, 60, 72, 84\}$.
The reason why this is so is since $\gcd(12,96) = \gcd(36,96) = \gcd(60,96) = \gcd(84,96) = 12$ but $\gcd(24,96) = 24$...