Combining symmetries into a group action

Let's combine some symmetries of the $4\times 4$ regular grid.

Translating (with wraparound) in the vertical direction is equivalent to a group action by $\mathbb{Z}_4$. Likewise in the horizontal direction. Since these mappings commute, we can combine them into a single group action by $\mathbb{Z}_4 \bigoplus \mathbb{Z}_4$, the translations.

But the $4\times 4$ regular grid also has the symmetries of a square, which represents the group action by dihedral group $D_8$ (many authors prefer the notation $D_4$), the reflections and rotations.

If we combine the group actions indicated by $\mathbb{Z}_4 \bigoplus \mathbb{Z}_4$ and $D_8$, of what larger group would that be the group action? Both kinds of group actions can be composed as permutations on the set of sixteen points in the grid, so the question makes sense as generating a smallish subgroup of $S_{16}$. But these actions do not commute in general, so the resulting subgroup should not be a direct sum of $\mathbb{Z}_4 \bigoplus \mathbb{Z}_4$ and $D_8$.

Is this an example of a semidirect product?

Did I sleep through the wrong class in abstract algebra?

Motivation: I want to search the 4-subsets of these sixteen points for a certain property, which I know is preserved by both kinds of symmetries. Rather than check all 1820 such subsets, it suffices to check only one representative in each orbit of the combined group action. It would be nice if one can generate an orbit by first applying the dihedral group action, followed by translation.

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This is indeed a semidirect product. The only thing that requires serious checking here is the fact that if you conjugate a translation by an element of $D_8$ (i.e. by a symmetry or a rotation), what you will get is another translation. In other words, $\mathbb{Z}_4 \times \mathbb{Z}_4$ is normalized by $D_8$ (if we interpret both as subgroups in $S_{16}$).
Thanks. An implication of this is that any combined group action can be expressed as a dihedral symmetry followed by a translation (because if $\tau ' = g\tau g^{-1}$ where $g \in D_8$ and $\tau,\tau '$ are translations, then $g\tau = \tau ' g$). –  hardmath Mar 9 '13 at 18:08