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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

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    $\begingroup$ It may help to exhibit each group as a product of groups $G_i$, where the $G_i$ are groups of prime power order. $\endgroup$ – André Nicolas Mar 31 '14 at 21:17
  • $\begingroup$ Recall that $\mathbb{Z}_m\times\mathbb{Z}_n$ is cyclic and $\cong \mathbb{Z}_{nm}$ iff $n$ and $m$ are relatively prime. Using this you can decompose your two groups to find that they are not isomorphic. See here for more info math.stackexchange.com/questions/354174/… $\endgroup$ – Sergio Da Silva Mar 31 '14 at 21:25
  • $\begingroup$ @SergioDaSilva Thanks for the link. In amWhy's response, why doesn't he decompose the given groups further (why does he leave one term as $(Z_8)^2$ in the first group)? $\endgroup$ – bclayman Mar 31 '14 at 21:29
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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.

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  • $\begingroup$ Thanks for the response, very helpful. So how do you know when to stop factorizing? I guess I'm not exactly sure how I'll know when I've arrived at the "canonical" form. Thanks again! $\endgroup$ – bclayman Mar 31 '14 at 21:49
  • $\begingroup$ @Mariogs If $(p^r,q)=1 then $Z_{p^rq}=Z_{p^r}\times Z_q$ so for each component of the product and each prime, you take the highest power you can - and then you don't split that up any more. I left brackets in to show where the components arise - look at how the powers of two are separated and gathered. $\endgroup$ – Mark Bennet Mar 31 '14 at 21:52
  • $\begingroup$ Hmm, so here's an example: how do you know to split $Z_24$ into 8 and 3 instead of 2 and 12, say? $\endgroup$ – bclayman Mar 31 '14 at 22:19
  • $\begingroup$ @Mariogs Because the highest power of $2$ which is a factor of $24$ is $8$ - $2$ and $12$ have the common factor $2$. $\endgroup$ – Mark Bennet Mar 31 '14 at 22:23
  • $\begingroup$ I see, and then you group elements with a common factor, e.g. (8 x 4 x 4) x 3 x 5, yeah? $\endgroup$ – bclayman Mar 31 '14 at 22:26

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