Show that $aI=bJ$ where $I,J$ are ideals in $\mathcal{O}_K=\mathbb{Z}[\sqrt{-6}]$ Let $K=\mathbb{Q}(\sqrt{-6})$, and therefore $\mathcal{O}_K=\mathbb{Z}[\sqrt{-6}]$. I have already proved that $[I]=[J]$, where $I=(2,\sqrt{-6})$ and $J=(3,\sqrt{-6})$. Now I want to  find $a,b \in \mathcal{O}_K$ so that $aI=bJ$.
Since $||I||=2$ and $||J||=3$, I know that $3|N(a)$ and $2|N(b)$. If I can find $A,B$ ideals so that $AI$ and $BJ$ are principal, then I have the desired result using franz lemmermeyer's method here. However, I am not able to do this. Can someone help me?
 A: The class group of a number field is finite, so all of the ideals have finite order. Thus some powers of $I$ and $J$ must be principal. In this case, it turns out that both have order $2$:
\begin{align*}
I^2 &= \left(2, \sqrt{-6}\right)^2 = \left(4, 2 \sqrt{-6}, -6\right) = \left(-2, 4, 2 \sqrt{-6}\right) = (2)\\
J^2 &= \left(3, \sqrt{-6}\right)^2 = \left(9, 3 \sqrt{-6}, -6\right) = \left(3, 9, 3 \sqrt{-6}\right) = (3) \, .
\end{align*}
Thus you can take $A = I$ and $B = J$ and apply the method from the linked post.
But we can actually find the desired $a$ and $b$ with a bit more work. The quantity $aI=bJ$ you're looking for is a multiple of both $I$ and $J$, hence is divisible by their lcm, which is $I \cap J$. Since $I$ and $J$ are comaximal ($1 = -2 + 3 \in I + J$), then $I \cap J = IJ$, and we compute
\begin{align*}
IJ &= \left(2, \sqrt{-6}\right) \, \left(3, \sqrt{-6}\right) = \left(6, 2\sqrt{-6}, 3\sqrt{-6}, -6\right) = \left(6, \sqrt{-6}\right) = \left(\sqrt{-6}\right) \, .
\end{align*}
Thus
\begin{align*}
\sqrt{-6} I = IJI = I^2 J = 2J
\end{align*}
as was stated in the comments, or similarly,
\begin{align*}
\sqrt{-6} J = IJJ = I J^2 = 3I \, .
\end{align*}

As a note, there is a general algorithm for computing the inverse of a fractional ideal of a number field $K$, assuming one has an integral basis for $\mathcal{O}_K$. This is given in $\S4.8.4$ (p. $202$) of Cohen's A Course In Computational Algebraic Number Theory.
