Groups of Order 16 with 4-Torsion, Up to Isomorphism

  • #1
WWGD
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Main Question or Discussion Point

Hi, I am trying to find all groups G of order 16 so that for every y in G, we have y+y+y+y=0.

My thought is using the structure theorem for finitely-generated PIDs. So I can find 3:

## \mathbb Z_4 \times \mathbb Z_4##,
## \mathbb Z_4 \times \mathbb Z_2 \times \mathbb Z_2 ## , and:

## \mathbb Z_2 ^4 ## .

How can I tell if these are the only 3 groups with this property up to isomorphism?

Thanks.
 

Answers and Replies

  • #2
lavinia
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You definitely have all of the abelian groups. All abelian groups are direct products of cyclic groups and these must all have order that divides 16.

But a non-abelian group of order 8 direct product ## \mathbb Z_2## will also work.
 
  • #3
WWGD
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What do you mean all Abelian groups are direct product of cyclic groups? What about ## \mathbb Z ## ?
 
  • #4
lavinia
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What do you mean all Abelian groups are direct product of cyclic groups? What about ## \mathbb Z ## ?
Not sure what you are asking here. You only need to consider finite abelian groups.

Look up the structure theorem for abelian groups. Any abstract algebra book should have it.
 
  • #5
WWGD
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Yes, I missed the obvious fact that ## \mathbb Z ## is cyclic. In my mind I was reading "torsion" , instead of cyclic.
 
  • #6
lavinia
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Yes, I missed the obvious fact that ## \mathbb Z ## is cyclic.
Still you only need finite cyclic groups for your problem.

In any case, the dihedral group of order 8, ## \mathbb D_8, ## is non-abelian.

## \mathbb D_8 \times \mathbb Z_2## is completely 4 torsion.

The relations for ## \mathbb D_8## are

## \ b^4 = a^2 ##

and ## \ aba = b^3##
 
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  • #7
WWGD
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Still you only need finite cyclic groups for your problem.

In any case, the dihedral group of order 8 is non-abelian.

## \mathbb D_8 \times \mathbb Z_2## is completely 4 torsion.

The relations for ## \mathbb D_8## are

## \mathbb b^4 = a^2 ##
Right, only finite cyclic groups are torsion. Also had forgotten this obvious fact too. Well, since there is only one infinite cyclic group up to isomorphism (right)? .
 
  • #8
lavinia
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Right, only finite cyclic groups are torsion. Also had forgotten this obvious fact too. Well, since there is only one infinite cyclic group up to isomorphism (right)? .
yes. Though you replied before I finished writing so my answer is incomplete. I hadn't finished the relations for the dihedral group of order 8. It is complete now.
 
  • #9
lavinia
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BTW: You might enjoy taking a hands on look at the dihedral group of order 8 in matrix form.

Let b be a rotation of the plane by 90 degrees counter clockwise and let a be reflection around the x-axis. These two symmetries of the plane generate a group. Try multiplying out all of the matrices.

The matrix for b is (1,0) -> (0,1) (0,1) -> (-1,0)
The matrix for a is (1,0) -> (1,0) (0,1) -> (0,-1)

Sorry I don't know how to write matrices properly here.

You can check directly that the fourth power of each matrix is the identity matrix.

Try coming up with other representations of ##\mathbb D_8##
 
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  • #10
jbunniii
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Any group of order 16 which does not contain an element of order 16 or 8 will meet the conditions.

In addition to the ones listed so far, there is ##Q \times \mathbb{Z}_2## (direct product of the quaternion group with ##\mathbb{Z}_2##).

There are also (non-direct) semidirect products of the form ##\mathbb{Z}_4 \rtimes \mathbb{Z}_4## and ##(\mathbb{Z}_2 \times \mathbb{Z}_2) \rtimes \mathbb{Z}_4##, but one has to do a bit of work to show that these exist. (And in general, they may not be unique; it's possible for two semidirect products with the same factors ##N \rtimes H## to be non-isomorphic.)
 
  • #11
lavinia
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As jbunniii wrote, there are other possible non-abelian groups of order 16 all of whose elements are 4 torsion.
And some of these may be viewed as semi-direct products of abelian groups.

One can create a semi-direct product of two groups H and A whenever A acts as a group of automorphisms of H.
The multiplication is defined on the set HxA by the rule

(h,a).(j,b) = (h(a.j),ab) where a.j denotes the action of a on the element,j.

This means that if one wants to find semi-direct products, one needs to find actions of A on H by a group of automorphisms.

So for the dihedral group of order 8, the non-zero element of ##\mathbb{Z}_2## acts on a generator of ##\mathbb{Z}_4## by inversion.
Similarly ##\mathbb{Z}_4## can act on ##\mathbb{Z}_4## by inversion. That is; the generator of one ##\mathbb{Z}_4## maps each element of the other to its inverse. This gives a group of order 16 all of whose elements are 4 torsion.

Question:

- What are the actions of

##\mathbb{Z}_4## on ##\mathbb{Z}_4## and on ## \mathbb{Z}_2 \times \mathbb{Z}_2 ## ?

Are the resulting semi-direct products isomorphic?

- Are there any actions of

##\mathbb{Z}_2## on ##\mathbb{Z}_4 \times \mathbb{Z}_2 ## ?
 
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