Are these two groups isomorphic? I'm looking at Section 11, problem 18 in Fraleigh.  Here's the question:
Is $Z_8 \times Z_{10} \times Z_{24}$ isomorphic to $Z_4 \times Z_{12} \times Z_{40}$?
I can do the problem once I figure classify each according to the fundamental theorem of finitely generated abelian groups.  (If my first group is isomorphic to X and so is my second group, then they're isomorphic to each other.)
How exactly do you decide what direct product of $Z_i$'s a given group is isomorphic to?
I would have said:
$8 * 10 * 24$ = $2^7 * 3 * 5$ = $4 * 12 * 40$, but this isn't enough to conclude my first group is isomorphic to:
$Z_{2^7}$ x $Z_3$ x $Z_5$, right?
Any help much appreciated,
Mariogs
 A: There are a couple of canonical forms for abelian groups. You have to be really careful just using the prime factorisation - for example $2^7=2\times 2\times 2\times 2\times 2\times 2\times 2$, but that doesn't mean $Z_{128}=Z_2\times Z_2\times Z_2\times Z_2\times Z_2\times Z_2\times Z_2$.
What you have to do is factorise more carefully.
Your first group presentation gives factors $8\times (2\times 5)\times (8\times 3)=(8\times 8\times 2)\times 3 \times 5$ when the prime factors are collected together. If you get the same prime components, you have the same group.
Another canonical form is $Z_{n_1}\times Z_{n_2}\times \dots Z_{n_r}$ where $r_{i+1}|r_i$
Here that comes out as $Z_{120}\times Z_8\times Z_2$. If two abelian groups have the same canonical form they are the same.
The second presentation gives $4\times(3\times 4)\times (8\times 5)=(8\times 4\times 4)\times 3 \times 5$ split into prime components or $Z_{120}\times Z_4\times Z_4$ in the canonical form.
This is different - to see this on a smaller scale note that $Z_8\times Z_2\neq Z_4\times Z_4$ - the first group here has an element of order $8$ while the second does not.
Although there are sometimes short cuts, reducing to an appropriate canonical form always works.
