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
How can non-isomorphic groups be isomorphic to the same group!
You can use abelian groups. But you do not have to.Any ideas? Do I start with the primary decomposition Z2^2 x Z3^4 x Z7
The dihedral group is one.
The cyclic group is another.
This group is another.
You can now use .
Aha, that makes more sense.
I presume that the A we're talking about is:
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 where is a factor of 126 and 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.
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.
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.