Published: Mon 29 October 2007
By Michi
In Algebra, Mathematics .

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 [tex]H \wr_X G[/tex] is defined for groups G,H
and a G-set X by the following data. The elements of [tex]H \wr_X
G[/tex] are tuples [tex](h_{x_1},h_{x_2},\dots,h_{x_r};g)\in
H^{|X|}\times G[/tex]. The trick is in the group product. We define

[tex](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')[/tex]

(or possibly with a lot of inverses sprinkled into those indices)

Consider, first, the case of [tex]G=H=X=C_2[/tex] with the nontrivial
G-action defined by gx=1, g1=x. We get 8 elements in the wreath product
[tex]H \wr_X G[/tex]. Thus, the group is one of the groups with 8
elements - [tex]C_8, C_4\times C_2, C_2^3, Q, D_4[/tex]. 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 ,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
[tex]C_8[/tex]. It has elements of order 4, and thus it's also not
[tex]C_2^3[/tex]. 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 [tex]C_4\times C_2[/tex].

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

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
[tex]x,y,z,a,b,c\in C_2[/tex], we get the multiplication

((x,y),z)*((a,b),c) = ((xa,yb),zc) - which is precisely the
multiplication in [tex](C_2\times C_2)\times C_2[/tex]. Thus,
this wreath product is just [tex]C_2^3[/tex].

Finally, for a more challenging group identification, we consider
[tex]G=X=C_3[/tex] with gx = x^{2} . Elements in the wreath
product [tex]H \wr_X G[/tex] have the form [tex]((x,y,z),w)\in
H^3\times G[/tex], 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: 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.