Let $\langle[3]_{12}\rangle$ be the normal subgroup of $\Bbb Z_{12}$ generated by $[3]_{12}$. Show that $$\Bbb Z_{12}/\langle[3]_{12}\rangle \cong \Bbb Z_p$$ for some $p \in \Bbb N$.
My hunch is that $p=11$ as the group $\Bbb Z_{12}/\langle[3]_{12}\rangle$ identifies $3$ with $0$ and thus we're left with $11$ elements.
I don't know what the map $f: \Bbb Z_{12}/\langle[3]_{12}\rangle \to \Bbb Z_{11}$ should be. If I take $a + [3]_{12} \longmapsto [a]_{11}$ I don't know wheter this satisfies the requirements?
The order of $H:=\langle [3]_{12}\rangle$ is four; this follows from $H$ being cyclic (as $G:=\Bbb Z_{12}$ is cyclic) and
$$\begin{align} 0[3]_{12}&=[0]_{12}, \\ 1[3]_{12}&=[3]_{12},\\ 2[3]_{12}&=[6]_{12},\\ 3[3]_{12}&=[9]_{12},\\ 4[3]_{12}&=[0]_{12}. \end{align}$$
Therefore, by Lagrange's Theorem, we have
$$\begin{align} \lvert G/H\rvert&=\lvert G\rvert/\lvert H\rvert\\ &=12/4\\ &=3. \end{align}$$
But the only group of order three is $\Bbb Z_3$ up to isomorphism.