Groups of Symmetries and presentation of Semi-direct products Let $L$ the tessellation of the euclidean plane by equilateral triangles. Prove that the group $G = Sym(L)$ of isometries of $ \mathbb{R}^2 $ leading $ L $ in $ L $ is is a semi-direct product $\mathbb{Z} \times \mathbb{Z} \rtimes D_6$, and obtain a presentation of this group.
 A: Hint/outline: Consider two subsets of the group of $\mathrm{Sym}(L)$, $T$ being the isometries that are pure translations, and $R$ the isometries that fix some $0\in L$.
Then you need to prove that $T$ and $R$ are subgroups of $\mathrm{Sym}(L)$, with $T\cong \mathbb Z\times \mathbb Z$ and $R\cong D_{12}$. (Here, I'm using $D_{12}$ as group of isometries on a hexagon, so it has $12$ elements.)
Then you need to prove every element of $\mathrm{Sym}(L)$ can be written as an isometry fixing $0$ follwed by a translation.  (That is, $\mathrm{Sym}(L)=TR$ or $RT$, depending on whether you write the group action on the left or right.)
Then you need to show that $T$ is a normal subgroup. The previous step means that you only need to show that $rtr^{-1}\in T$ when $r\in R$ and $t\in T$.  That's the hard trick, and it is easier to do for two generators of $R$.
All together, this would prove that $\mathrm{Sym}(L)\cong (\mathbb Z\times\mathbb Z)\rtimes D_{12}$.
Not sure if this can be used to prove that it isn't $\mathbb Z\times \mathbb Z\rtimes D_6$ - it might be possible to factor the group differently.
The "obvious" way to include $D_6$ in $\mathrm{Sym}(L)$ is to consider some atomic triangle, $a,b,c\in L$, and consider the isometries that send $\{a,b,c\}$ to itself.
However, then the other subgroup has to be something more complex than translations, since we need to be able to send $\{a,b,c\}$ to any triangle in $\mathrm{L}$, which can't happen since translation can send a triangle to only triangles oriented similarly. 
