Find all normal subgroups of $S_3 \times \mathbb{Z}_6$

Question

I know that if $H_1,H_2$ are normal subgroups of $G_1,G_2$, respectively, then $H_1\times H_2$ is a normal subgroup of $G_1\times G_2$. Since $\mathbb{Z}_6$ is abelian, all its 4 subgroups are normal. The 3 normal subgroups of $S_3$ are $S_3$, $A_3$ and the trivial subgroup. Thus I can get 12 normal subgroups of $S_3 \times \mathbb{Z}_6$, but I do not know whether there are others, or how to find them.

Answer 1

In addition to the 12 normal subgroups mentioned in the initial post, there are two other normal subgroups: $$ A_1=\{(e,\bar0),\,((12),\bar3),\,((13),\bar3),\,((23),\bar3),\,((123),\bar0),\,((132),\bar0)\}; $$ $$ A_2=\langle((123),\bar0),\,(e,\bar2),((12),\bar3)\rangle. $$ The subgroup $A_1$ is of order 6, and the subgroup $A_2$ contains 18 elements.