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

Gold Member
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.

Gold Member
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.

Gold Member
What do you mean all Abelian groups are direct product of cyclic groups? What about ## \mathbb Z ## ?

Gold Member
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.

Gold Member
Yes, I missed the obvious fact that ## \mathbb Z ## is cyclic. In my mind I was reading "torsion" , instead of cyclic.

Gold Member
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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Gold Member
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)? .

Gold Member
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.

Gold Member
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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Homework Helper
Gold Member
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.)

Gold Member
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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