How would I be able to tell if two direct products of cyclic groups are isomorphic to one another? I'm working with two pieces of knowledge so far:

1) The order of the direct product is the product of the orders of all the groups that make up the direct product 2) A direct product of cyclic groups $C_m$ and $C_n$ is only cyclic is $(m,n)=1$.

Are there any other techniques I can use? For example, I'm trying to determine is $C_4 X C_25$ is isomorphic to $C_{100}$.


The main technique is the Chinese remainder theorem: $C_m \times C_n \cong C_{mn}$ when $\gcd(m,n)=1$, which follows directly from your two pieces of knowledge.

This gives at once that $C_4 \times C_{25}$ is isomorphic to $C_{100}$.

Using the Chinese remainder theorem repeatedly, you can decompose any direct product of several cyclic groups into a direct product of cyclic groups of prime-power order ($p$-groups); this is called the primary decomposition.

From the primary decomposition, you can reassemble the factors into the form $C_{d_1} \times C_{d_1} \times \cdots \times C_{d_n}$ with $d_1 \mid d_2 \mid \cdots \mid d_n$; this called the invariant factor decomposition.

Both forms are canonical forms and two direct product of cyclic groups are isomorphic iff they have the same canonical forms.

All this is the easy part of the structure theorems for finite abelian groups. The hard part is proving that every finite abelian group is a direct product of cyclic groups. See Simple proof of the structure theorems for finite abelian groups. Uniqueness also takes some work.

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  • $\begingroup$ So for example, $C_{100}$ will decompose into $C_2^2 \times C_5^2$, and $C_{10} \times C_{10}$ will decompose into $C_2 \times C_5 \times C_2 \times C_5$, so the two are isomorphic? $\endgroup$ – jstnchng Oct 29 '14 at 2:56
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    $\begingroup$ @JC1397, yes. But in this case, it's easier, because $C_{100}$ is cyclic of order $100$ but in $C_{10} \times C_{10}$ every element has order at most $10$, and so they can't be isomorphic. $\endgroup$ – lhf Oct 29 '14 at 2:58
  • $\begingroup$ But don't they have the same direct products? $\endgroup$ – jstnchng Oct 29 '14 at 3:22
  • $\begingroup$ @JC1397, no, because $C_4$ is not isomorphic to $C_2\times C_2$, for instance. $\endgroup$ – lhf Oct 29 '14 at 3:28
  • $\begingroup$ So what is the direct product of $C_4$? How would I decompose $C_10 \times C_10$? $\endgroup$ – jstnchng Oct 29 '14 at 3:29

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