Wreath products

Published: Mon 29 October 2007

In a conversation on IRC, I started prodding at low-order wreath products. It turned out to be quite a lot of fun doing it, so I thought I'd try to expand it into a blog post.

The wreath product $$H \wr_X G$$ is defined for groups G,H and a G-set X by the following data. The elements of $$H \wr_X G$$ are tuples $$(h_{x_1},h_{x_2},\dots,h_{x_r};g)\in H^{|X|}\times G$$. The trick is in the group product. We define
$$(h_{x_1},h_{x_2},\dots,h_{x_r};g)\cdot (h'_{x_1},h'_{x_2},\dots,h'_{x_r};g')= \\ (h_{x_1}h'_{gx_1},h_{x_2}h'_{gx_2},\dots,h_{x_r}h'_{gx_r};gg')$$
(or possibly with a lot of inverses sprinkled into those indices)

Consider, first, the case of $$G=H=X=C_2$$ with the nontrivial G-action defined by gx=1, g1=x. We get 8 elements in the wreath product $$H \wr_X G$$. Thus, the group is one of the groups with 8 elements - $$C_8, C_4\times C_2, C_2^3, Q, D_4$$. We shall try to identify the group in question using orders of elements as the primary way of recognizing things. Consider an element ((x,y),z).

If z=1, then this squares to ((x:sup:2,y2),1) = ((1,1),1), so all these elements have order 2 (except for ((1,1),1) which has order 1).

If z=g, then the element squares to ((xy,yx),1). So if x=y, then the element squares to the identity, and thus has order 2. Remains two elements where this is nontrivial - namely ((1,h),g) and ((h,1),g) - both of which square to ((h,h),1), which squares to the identity. And thus these two elements have order 4.

Thus, no element of order 8 - and thus the group is not $$C_8$$. It has elements of order 4, and thus it's also not $$C_2^3$$. A reasonable question at this point is whether it's abelian. Consider the following products.
((1,h),g) * ((1,h),1) = ((h,h),g)
((1,h),1) * ((1,h),g) = ((1,1),g)
thus eliminating $$C_4\times C_2$$.

This leaves $$Q,D_4$$. At this point, I'm basically going to make a lucky guess and give an isomorphism $$H\wr_X G = D_4$$. Let $$D_4=\langle a,b\mid a^4=b^2=abab=1\rangle$$.

We send ((1,h),g) = a, giving ((h,h),1) = a2 and ((h,1),g) = a3 = a-1. We then pick any one of the remaining nontrivial elements, say ((1,h),1) = b. We need to verify the relations in the group presentation. a4=1 and b2=1 are already reasonably obvious. So we compute
((1,h),g)*((1,h),1)*((1,h),g)*((1,h),1) =
((h,h),g)*((h,h),g) = ((1,1),1). Which concludes the identification.
For a second take, we consider X to have trivial G-action. Then, given $$x,y,z,a,b,c\in C_2$$, we get the multiplication
((x,y),z)*((a,b),c) = ((xa,yb),zc) - which is precisely the multiplication in $$(C_2\times C_2)\times C_2$$. Thus, this wreath product is just $$C_2^3$$.
Finally, for a more challenging group identification, we consider $$G=X=C_3$$ with gx = x2. Elements in the wreath product $$H \wr_X G$$ have the form $$((x,y,z),w)\in H^3\times G$$, and we get the multiplication as
((x,y,z),w)*((a,b,c),d) = ((x,y,z)w(a,b,c),wd).
The group will be of order 24, so there's a few more to choose from. For an elimination of all the abelian groups, consider
((h,1,1),g) * ((1,h,1),g) = ((h,h,1),g:sup:2)
((1,h,1),g) * ((h,1,1),g) = ((1,1,1),g:sup:2)
Now, all elements ((x,y,z),1) have order 2, so to maximize order, we need something like ((x,y,z),g). Now consider the powers of this element.
((x,y,z),g)
((xz,yx,zy),g:sup:2)
((xzy,yxz,zyx),1) - which if exactly two of x,y,z are non-identity vanishes, and otherwise is nontrivial.
((xzyx,yxzy,zyxz),g)
((xzyxz,yxzyx,zyxzy),g:sup:2)
((xzyxzy,yxzyxz,zyxzyx),1) = ((1,1,1),1)

So, no element has order more than 6. This, again, rules out many of the candidate groups.

To be specific - we can now partition the elements after order.
((1,1,1),1) - 1 element, order 1
((x,y,z),1) - 7 elements, order 2
((1,1,1),g), ((h,h,1),g), ((h,1,h),g), ((1,h,h),g),
((1,1,1),g:sup:2), ((h,h,1),g:sup:2), ((h,1,h),g:sup:2), ((1,h,h),g:sup:2) - 8 elements, order 3
((h,1,1),g), ((1,h,1),g), ((1,1,h),g), ((h,h,h),g),
((h,1,1),g:sup:2), ((1,h,1),g:sup:2), ((1,1,h),g:sup:2), ((h,h,h),g:sup:2) - 8 elements, order 6
At this point, I grab Google as a research assistant. It turns up 15 groups of order 24. Using the small groups database enumeration, we find:
24.1: has an element of order 12.
24.2: has an element of order 24. Also is abelian.
24.3: has an element of order 4.
24.4: has an element of order 4.
24.5: has an element of order 4.
24.6: has an element of order 4.
24.7: has an element of order 4.
24.8: has an element of order 4.
24.9: is abelian.
24.10: has an element of order 4.
24.11: has an element of order 4.
24.12: has an element of order 4.
24.13: A4xC2
24.14: D6xC2
24.15: is abelian.

Now, with the candidate groups narrowed down this far, we consider the two remaining groups, and (in my case, using Magma), and grab the number of elements of each order.

A4xC2:
1 element of order 1
7 elements of order 2
8 elements of order 3
8 elements of order 6
D6xC2:
1 element of order 1
15 elements of order 2
2 elements of order 3
6 elements of order 6

which seals the deal. The group we've constructed is A4xC2.

Exhibiting an explicit isomorphism goes beyond what I feel like doing right this evening, and leave it as a nice exercise for the interested reader.