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.
First off, we’ll start with a definition:
The wreath product is defined for groups G,H and a G-set X by the following data. The elements of are tuples . The trick is in the group product. We define
(or possibly with a lot of inverses sprinkled into those indices)
Consider, first, the case of with the nontrivial G-action defined by gx=1, g1=x. We get 8 elements in the wreath product . Thus, the group is one of the groups with 8 elements – . 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 ((x2,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 . It has elements of order 4, and thus it’s also not . 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 .
This leaves . At this point, I’m basically going to make a lucky guess and give an isomorphism . Let .
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
((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 , we get the multiplication
((x,y),z)*((a,b),c) = ((xa,yb),zc) – which is precisely the multiplication in . Thus, this wreath product is just .
Finally, for a more challenging group identification, we consider with gx = x2. Elements in the wreath product have the form , 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),g2)
((1,h,1),g) * ((h,1,1),g) = ((1,1,1),g2)
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.
((xzy,yxz,zyx),1) – which if exactly two of x,y,z are non-identity vanishes, and otherwise is nontrivial.
((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),g2), ((h,h,1),g2), ((h,1,h),g2), ((1,h,h),g2) – 8 elements, order 3
((h,1,1),g), ((1,h,1),g), ((1,1,h),g), ((h,h,h),g),
((h,1,1),g2), ((1,h,1),g2), ((1,1,h),g2), ((h,h,h),g2) – 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.
24.15: 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.
1 element of order 1
7 elements of order 2
8 elements of order 3
8 elements of order 6
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.