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The problem is to find a presentation of the regular wreath product of $C_4$ and $D_6$ (the dihedral group of order 6).

Note: By definition the wreath product is a semidirect product $(C_4)^6 \rtimes D_6$ realised by a specific homomorphism $ \theta$, one should inculude the relations of the free product $(C_4)^6 * D_6$:
{$a^2=...=f^2=1, [a,b]=[a,c]=...=[e,f]=1, x^6=y^2=1, xy=yx^-1$} besides one needs adding the conjugates $a^x,a^y, b^x,b^y...$.

My query: What tuple should $a$ be taken as in order to compute $a^x=\theta_x (a)$? For example given $a$ is the generator of the first (left-most) copy of $C_4$ in $(C_4)^6$ then is $\theta_x (a)=\theta_x((1,0,0,0,0,0))$?

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Indeed the $\theta$ depends on an pre-chosen ordering of elements of H, but I just not sure what tuple to be taken for a in order to calculate $\theta_x(a)$ –  user31899 May 31 '12 at 9:43

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up vote 2 down vote accepted

The notation $D_6$ is (unfortunately) sometimes used to mean the dihedral group of order 6 and sometimes the dihedral group of order 12. It does not help us to decide which you mean if you first refer to it as the dihedral group of order 6, but then proceed to write down relations $x^6=y^2=1, xy=yx^{-1}$ for the dihedral group of order 12.

But let's assume you mean dihedral of order 6, so you should have $x^3=1$, not $x^6=1$. You should start by writing down the action of $x$ and $y$ in the regular permutation representation of $D_6$; for example $x \to (1,2,3)(4,5,6)$, $y \to (1,6)(2,5)(3,4)$. You can then calculate the conjugation relations: $a^x=b,b^x=c,c^x=a,d^x=e$, etc, $a^y=f,f^y=a$, etc.

Incidentally, it is easy to see that many of your relations are redundant. By omitting obviously redundant relations, we get the presentation:

$\langle a,b,c,d,e,f,x,y \mid$

$\ \ \ \ a^4= [a,b]=[a,c]=[a,d]=[a,e]=[a,f]=x^3=y^2=(xy)^2=1,$

$\ \ \ \ a^x=b,b^x=c,d^x=e,e^x=f,a^y=f,b^y=e,c^y=d \rangle$.

Added later: The final two relators $b^y=e,c^y=d$ are also redundant, because they follow from the relations of $D_6$ and the other conjugation relations!

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