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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.

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What have you tried? –  Tobias Kildetoft Feb 13 '13 at 20:54
    
$D_6$ denote the dihedral group with six elements –  Agenor Andrade Feb 13 '13 at 20:55
    
Do you know how to characterize the group of all isometries of the Euclidean plane as a semi-direct product? –  Thomas Andrews Feb 13 '13 at 21:13
    
I'm not sure you are correct about which $D_6$ you want - it seems like it ought to be the group of symmetries of a regular hexagon, which means $D_6$ has $12$ elements. Maybe I'm missing something. –  Thomas Andrews Feb 13 '13 at 21:19
    
Thomas Andrews I do not know how characterize the group of all isometries of the Euclidean plane as semi-direct product. And the notation for $D_6$ is the dihedral group with six elements ($D_{2.3} = D_6$). Thanks. –  Agenor Andrade Feb 13 '13 at 21:29

1 Answer 1

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.

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I don't see how it could be isomorphic to $({\mathbb Z} \times {\mathbb Z}) \rtimes D_6$, because no group with that structure could contain an element of order 6. –  Derek Holt Feb 14 '13 at 8:26
    
Yeah, that was my feeling, too, @DerekHolt, but when I tried to prove that, I hit stumbling blocks. –  Thomas Andrews Feb 14 '13 at 13:08
    
Where is the stumbling block? –  Derek Holt Feb 14 '13 at 15:01
    
Ah, I was just being brain-dead. Yeah, it is fairly easy to show. –  Thomas Andrews Feb 14 '13 at 15:20
    
Forgive my ignorance gentlemen, @ThomasAndrews but I can not prove that $Sym(L) $ can be written to an isometry fixing 0 follwed by a translation. Nor can show that $ T $ is $ \mathbb{Z} \times \mathbb{Z} $ and $ R $ is $ D_{12} $ .. Would greatly help if you could show me more this way. thank you –  Agenor Andrade Feb 14 '13 at 19:58

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