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 ,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
[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) = 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
[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 = x2 . 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: A4 xC2
24.14: D6 xC2
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.
A4 xC2 :
1 element of order 1
7 elements of order 2
8 elements of order 3
8 elements of order 6
D6 xC2 :
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
A4 xC2 .
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.