Non isomorphic groups

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July 19th 2008, 12:22 PM
moolimanj

Non isomorphic groups

Hi

I need to find four non-isomorphic groups of order 2268 (expressed as products of their p-primary components), each isomorphic to A=Z126 x Z108.

Any ideas? Do I start with the primary decomposition Z2^2 x Z3^4 x Z7
July 19th 2008, 06:35 PM
ThePerfectHacker

Quote:

Originally Posted by moolimanj

Hi

I need to find four non-isomorphic groups of order 2268 (expressed as products of their p-primary components), each isomorphic to A=Z126 x Z108.

How can non-isomorphic groups be isomorphic to the same group!

Quote:

Any ideas? Do I start with the primary decomposition Z2^2 x Z3^4 x Z7

You can use abelian groups. But you do not have to.
The dihedral group $D_{1134}$ is one.
The cyclic group $\mathbb{Z}_{2268}$ is another.
This group $S_3\times \mathbb{Z}_{378}$ is another.
You can now use $\mathbb{Z}_2^3 \times \mathbb{Z}_3^4 \times \mathbb{Z}_7$ .
July 20th 2008, 06:41 AM
moolimanj

1 Attachment(s)

Thanks, but I dont understand your working. Perhaps I have interpreted the question wrong - can you explain a bit more. I have attached the original question:
July 20th 2008, 07:37 AM
Matt Westwood

Aha, that makes more sense.

I presume that the A we're talking about is:

$<br /> A = \mathbb{Z}_{126} \times \mathbb{Z}_{108}<br />$

So right, the way I would attack this would be to factor 126 and 108 and work out the ways you can express 2268 in such a way as for it to be $x \times y$ where $x$ is a factor of 126 and $y$ is a factor of 108.

But it's a few years since I did M336, maybe someone else can come up with a fuller answer than this.
July 20th 2008, 08:02 AM
moolimanj

Well, 2268 can be expressed in the following ways:

Z2268
Z2 X Z1134
Z3 X Z756
Z6 X Z378
Z9 X Z252
Z18 X Z126
Z3 X Z3 X Z252
Z3 X Z6 X Z126
Z3 X Z3 X Z3 X Z84
Z3 X Z3 X Z6 X Z42

But ow do i identify the non-isomorphic groups? Whats the method
July 20th 2008, 08:20 AM
Matt Westwood

Can't remember the details, but because Z126 and Z108 are not coprime, their product can not be a cyclic group. It's there in the notes somewhere.

So if you factor Z126 and Z108 and work out the bits of it that *are* coprime, you can work out the maximum order of a cyclic group that A can have, and you'll probably find that 268 is not a factor of it.
July 20th 2008, 08:26 AM
moolimanj

I still cant see the answer.

for Z126 I get Z2 x Z3^2 x Z7
and for Z108, I get Z2^2 x Z3^2

Thus A = (Z2 x Z2^2) x (Z3^2 x Z3^3) x Z7

Any ideas?
July 20th 2008, 10:21 AM
Matt Westwood

Z2268 is a cyclic group. So by one of the results you'll have encountered already, all its subgroups are cyclic.

Z126 = Z2 x Z3^2 x Z7 (all three of these are coprime).

Z108 = Z2^2 x Z3^3 (both of these are coprime)

To get Z2268 you need factors Z2^2 x Z3^4 x Z7.

Z2268 needs a factor Z3^4 = Z81 and you can't get that by combining factors of Z126 and Z108, because, say, Z3 x Z3^3 is not a cyclic group. Zm x Zn is cyclic only when m and n are coprime.

Study that part of your module again (see what I said last posting) and you'll find what I'm talking about.