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Geometric understanding of Semi-direct product

  1. Jul 7, 2013 #1
    In undergraduate abstract algebra we are not exposed to semi-direct products, so I was hoping someone could help me as I am doing some research in this area.

    I am familiar with the definitions of direct products and normal groups, and I know that a semidirect product is one where one of the groups of the product is not normal. But I don't have a good intuitive understanding of what his means. To put it in more blunt terms, why should I care whether a group is normal or not?

    I am working with the group of isometries in the taxicab metric, which is a semidirect product of T(2) and D4. T(2) is the normal group in this case.

    But let's start with something simpler. The Euclidean group of isometries of the plane is a semidirect product of O(2) and T(2). So all reflections, rotations, and transformations can be represented by combining the elements of these two groups. What does T(2) being "normal" to this group mean, in the geometric sense?

    I hope my question makes sense. If not I will try and clarify.

    -Dave K
  2. jcsd
  3. Jul 7, 2013 #2


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    In a direct product elements of the two groups commute with each other. This means that as transformations, order is irrelevant.

    In a semidirect product, order matters.

    But there is more.

    Formally G is the semi-direct product of K and H if there is a split exact sequence

    1 -> K -> G -> H ->1

    This means two things:

    - K is a normal subgroup of G such that G/K is isomorphic to H.

    - H lifts to a subgroup of G.

    Normal means that H acts on K as a group of automorphisms through conjugation.
    The action of H on K represents the deviation from the direct product. If the action is trivial G is in fact a direct product.

    Here is a simple example. The subgroup of O(2) generated by a 90 degree rotation and a reflection about the y-axis.

    The rotation and reflection do not commute so the order in which they are performed leads to different transformations of the plane.

    This group is the semi-direct product

    0 -> Z[itex]_{4}[/itex] -> G -> Z[itex]_{2}[/itex] ->0

    - Z[itex]_{4}[/itex] is the group generated by the 90 degree rotation.
    - Z[itex]_{2}[/itex] is the subgroup generated by the reflection.

    In general an exact sequence of groups may not describe a semi-direct product because there may not be a lift of H to G.

    For instance the exact sequence

    0 -> Z[itex]_{2}[/itex] -> Z[itex]_{4}[/itex] -> Z[itex]_{2}[/itex] -> 0

    is not a semidirect product.
    Last edited: Jul 8, 2013
  4. Jul 9, 2013 #3
    You care about normal groups because it allows us to simpify the group. Indeed, if we have a group ##G## with a normal subgroup ##N##, then we can study ##G## by studying the simpler groups ##N## and ##G/N##. We can keep going, and eventually we arive at groups with no normal subgroup. They are called simple groups. They can be seen as the basic building blocks of all groups.

    Very often, if we have a normal subgroup, then we can decompose ##G## as the semidirect product of two subgroups. In this case, the structure of ##G## is known. For example, if you want to find all groups of order 6, then you see that there are at most two possibilities. By explicitely constructing the groups ##\mathbb{Z}_2\times \mathbb{Z}_3## and ##\mathbb{Z}_3\langle \mathbb{Z}_2##, we obtain all the groups.

    A semi-direct product is given by some kind of action ##\alpha:H\rightarrow Aut(N)##. This action should be seen as the conjugation action. So what we do in general is constructing some kind of product of G and N with a specified conjugation action.
  5. Jul 10, 2013 #4


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    Here is another simple example with a different geometric picture.

    The direct product of the integers with itself, ZxZ, acts on the plane by translations as the lattice of vectors with integer coordinates. Specifically,

    (1,0).(x,y) = (x+1,y) and (0,1).(x,y) = (x,y+1)

    The quotient space of this action is a flat torus as can be seen by looking at the identifications on the unit square in the plane.

    The torus is the Cartesian product of two circles, S[itex]^{1}[/itex]xS[itex]^{1}[/itex] where the first factor is the circle arising from the x-axis in the plane and the second factor is the circles arising from the yaxis. The direct product is reflected in the symmetry of the torus i.e that interchanging the x and y axes projects to an isometry of the torus.

    That is:the interchange, (x,y) -> (y,x) commutes with the action of the integer lattice and so projects to a congruence of the torus.

    Algebraically, the direct product, ZxZ, can be described as a split extension of Z by Z

    0 -> Z -> ZxZ -> Z -> 0

    where the action of the right hand Z on the left hand Z is trivial.

    Now add to this lattice one more Euclidean motion, (x,y) -> (x+1/2, -y).

    The square of this motion is (x,y) -> (x + 1,y) so its square together with (x,y) -> (x,y+1) gives you the lattice back again.

    The group if Euclidean motions that this new motion together with (x,y) -> (x,y+1) projects to a flat Klein bottle and is also a split extension of Z by Z

    0 -> Z -> ZxZ -> Z -> 0

    but in this case the action of the right hand Z on the left hand is not trivial. It is multiplication by minus one.

    So this is a non-trivial semi-direct product of Z with Z.

    This non-triviality, the multiplication by -1, is reflected in the twisting of the Klein bottle. Interchanging the x and y axes non-longer projects to a congruence.
  6. Jul 10, 2013 #5
    Ok, this is helpful. I'll respond in more detail later, but wanted to express my appreciation to Lavinia and micromass.
  7. Jul 11, 2013 #6
    So, generally speaking, if I take any two groups, say H and K, then take H##\times##K, I will end up with a group (say G) in which both H and K are normal. Is that correct?

    But I cannot construct a semidirect product in this way - I would have to start with a Group G and be able to FIND two subgroups such that their product forms a semi-direct group. Is this generally how it's done?

    Please note that I am only now taking undergraduate abstract algebra, and the research I'm doing is a bit advanced for me, so I'm learning this in a top down fashion, starting from the larger concept of semi-direct groups and working my way down. Please forgive any lack of mathematical fluency in my asking these questions.

    -Dave K
  8. Jul 11, 2013 #7
    Correct. The converse is true too. That is, if you have a group with two normal subgroups ##N## and ##M## such that ##N\cap M=\{e\}## and ##NM=G##, then ##G##is (isomorphic to) the direct product ##N\times M##.

    Yes. And a generalization of the above theorem holds. Let's say that you have a group ##G## with two subgroups ##N## and ##H## such that ##N## is normal and ##N\cap H=\{e\}## and ##NH=G##, then ##G## is a semidirect product of ##N## and ##H##.

    So in general, you need to find TWO subgroups. However, there is a pretty cool theorem, called the Schur-Zassenaus theorem. This allows you to conclude the group is a semidirect product by only finding one normal subgroup which satisfies some hypothesis on its order: http://en.wikipedia.org/wiki/Schur–Zassenhaus_theorem
  9. Jul 11, 2013 #8
    This is more helpful (and timely) then you might realize, Micromass. More later...
  10. Jul 11, 2013 #9


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    If you have two groups then you can form a semi direct product given any action of one of the groups on the other.

    For instance, start with Z/4Z and Z/2Z let Z/2Z act on Z/4Z by the automorphism a -> a[itex]^{3}[/itex]

    Then the resulting group is the dihedral group of order 8

    It satifies the split exact sequence

    0 -> Z/4Z -> G -> Z/2Z -> 0

    Z/4Z is the normal subgroup and Z/2Z is isomorphic to the quotient group.

    OR Start with Z/2Z x Z/2Z and Z/2Z and let Z/2Z act on Z/2Z x Z/2Z by (a,b) -> (b,a).

    One then gets a semi direct product which is also isomorphic to the dihedral group or order 8.

    It satisfies the exact sequence,

    0 -> Z/2Z x Z/2Z -> G -> Z/2Z -> 0
    Last edited: Jul 11, 2013
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