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

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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.

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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.

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

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.

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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.

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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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)? .

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.

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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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.)

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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