Prime factorization of order and mutual non-isomorphism

Forum for the GRE subject test in mathematics.
hopeful_statguy
Posts: 7
Joined: Sun Sep 13, 2009 5:45 pm

Prime factorization of order and mutual non-isomorphism

Postby hopeful_statguy » Sun Sep 20, 2009 2:43 pm

I'm trying to understand why, in Example 6.7 (pg 239) of Cracking the GRE Math, we know that each unique list of elementary divisors gives rise to an abelian group that is non-isomorphic to the other groups defined by the list.
The groups are certainly the same order...
I understand that 6) Z_8 \oplus Z_3 \oplus Z_{25} is a black sheep because it is the only cyclic group (gcd of the m's is 1).
But what about the others? How do we know that they are mutually non-isomorphic?

btw, to write the above in tex, alls I typed was Z_8 \oplus Z_3 \oplus Z_{25}

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diogenes
Posts: 73
Joined: Sun Aug 31, 2008 9:31 pm

Re: Prime factorization of order and mutual non-isomorphism

Postby diogenes » Mon Sep 21, 2009 10:49 pm

I think one way you might see this is by attempting to express the groups in different ways by grouping them together if they are relatively prime, e.g.,:

Z_2 \oplus Z_2 \oplus Z_2 \oplus Z_3 \oplus Z_5 \oplus Z_5 \oplus \approx Z_6 \oplus Z_{10} \oplus Z_{10}

Now, looking at the remaining groups this way you will see that you cannot factor them into an isomorphic group.

Or, we could find the lcm for each of the groups (30,150,60,300,120,600) to find the largest possible order of an element in each group and starting sorting things out that way.

Also, this might be handy:

http://en.wikipedia.org/wiki/Finitely_g ... lian_group




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