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 ((x^{2},y^{2}),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) = a^{2} and ((h,1),g) = a^{3} = 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. a^{4}=1 and b^{2}=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 , 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 = x^{2}. 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),g^{2})

((1,h,1),g) * ((h,1,1),g) = ((1,1,1),g^{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^{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^{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^{2}), ((h,h,1),g^{2}), ((h,1,h),g^{2}), ((1,h,h),g^{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^{2}), ((1,h,1),g^{2}), ((1,1,h),g^{2}), ((h,h,h),g^{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: A_{4}xC_{2}

24.14: D_{6}xC_{2}

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.

A_{4}xC_{2}:

1 element of order 1

7 elements of order 2

8 elements of order 3

8 elements of order 6

D_{6}xC_{2}:

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 A_{4}xC_{2}.

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.

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