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

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- Jul 19th 2008, 12:22 PMmoolimanjNon 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 - Jul 19th 2008, 06:35 PMThePerfectHacker
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

The dihedral group $\displaystyle D_{1134}$ is one.

The cyclic group $\displaystyle \mathbb{Z}_{2268}$ is another.

This group $\displaystyle S_3\times \mathbb{Z}_{378}$ is another.

You can now use $\displaystyle \mathbb{Z}_2^3 \times \mathbb{Z}_3^4 \times \mathbb{Z}_7$. - Jul 20th 2008, 06:41 AMmoolimanj
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:

- Jul 20th 2008, 07:37 AMMatt Westwood
Aha, that makes more sense.

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

$\displaystyle

A = \mathbb{Z}_{126} \times \mathbb{Z}_{108}

$

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 $\displaystyle x \times y$ where $\displaystyle x$ is a factor of 126 and $\displaystyle y$ 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. - Jul 20th 2008, 08:02 AMmoolimanj
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 - Jul 20th 2008, 08:20 AMMatt 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. - Jul 20th 2008, 08:26 AMmoolimanj
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? - Jul 20th 2008, 10:21 AMMatt 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.