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Is the direct sum of cyclic p-groups a cyclic group?

by lugita15
Tags: cyclic, pgroups
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lugita15
#1
Feb29-12, 12:23 AM
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For arbitrary natural numbers a and b, I don't think the direct sum of Z_a and Z_b (considered as additive groups) is isomorphic to Z_ab. But I think if p and q are distinct primes, the direct sum of Z_p^m and Z_q^n is always isomorphic to Z_(p^m * q^n). Am I right? I've been freely using these facts in abstract algebra, but I wanted to make sure they're correct.

Any help would be greatly appreciated.

Thank You in Advance.
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Office_Shredder
#2
Feb29-12, 01:31 AM
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It's fairly simple to check.... if it's cyclic, the generator has got to be (1,1) (because seriously, what else would it be). Does that element have order pmqn?

Alternatively, this is an example of the chinese remainder theorem
http://en.wikipedia.org/wiki/Chinese..._ideal_domains

where R is the integers, u = pmqn
lugita15
#3
Feb29-12, 02:36 AM
P: 1,583
Thanks Office_Shredder!

Deveno
#4
Feb29-12, 06:09 AM
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Is the direct sum of cyclic p-groups a cyclic group?

as a counter-example Z2 x Z2 is not cyclic, as it has no element of order 4:

(0,0) is of order 1
(1,0) + (1,0) = (0,0)
(0,1) + (0,1) = (0,0)
(1,1) + (1,1) = (0,0)

so all other elements are of order 2.

in fact, it is not hard to show that if G is abelian (and AxB is abelian if A and B are, which is certainly true if A and B are cyclic) that

|xy| ≤ lcm(|x|,|y|), so if gcd(m,n) ≠ 1, then Zm x Zn cannot possibly be isomorphic to Zmn, since there aren't any elements of order mn.

on the other hand, the CRT is equivalent to saying

k → (k (mod m), k (mod n)) is an isomorphism of Zmn with Zm x Zn when gcd(m,n) = 1. this is clearly a homomorphism, so showing it's surjective is the hard part (which is really the same thing as showing (1,1) generates the direct product).

(actually the CRT usually gives the inverse isomorphism, and the construction of the solution to:

x = a mod m
x = b mod n

actually gives us the inverse isomorphism, which is the pre-image of the isomorphism above of (a,b):

x = an[n-1]m + bm[m-1]n (mod mn),

where [n-1]m denotes the inverse of n (mod m), which exists only when gcd(m,n) = 1).

i always knew that that "least common multiple" stuff they made me suffer through in grade school while doing fractions would pay off someday.
micromass
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Feb29-12, 09:34 AM
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Quote Quote by lugita15 View Post
For arbitrary natural numbers a and b, I don't think the direct sum of Z_a and Z_b (considered as additive groups) is isomorphic to Z_ab. But I think if p and q are distinct primes, the direct sum of Z_p^m and Z_q^n is always isomorphic to Z_(p^m * q^n). Am I right? I've been freely using these facts in abstract algebra, but I wanted to make sure they're correct.

Any help would be greatly appreciated.

Thank You in Advance.
Even something more general is true!!
Indeed, if gcd(a,b)=1, then [itex]\mathbb{Z}_{ab}\cong \mathbb{Z}_a\times \mathbb{Z}_b[/itex]. So the result is not only true for p-groups.
Deveno
#6
Feb29-12, 09:58 AM
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Quote Quote by micromass View Post
Even something more general is true!!
Indeed, if gcd(a,b)=1, then [itex]\mathbb{Z}_{ab}\cong \mathbb{Z}_a\times \mathbb{Z}_b[/itex]. So the result is not only true for p-groups.
and something even more general is true: every abelian group can be decomposed into a direct sum of cyclic groups, which in turn can be decomposed into cyclic p-groups (different p's, of course).

abelian groups are the well-mannered groups that always say please and thank-you. as such, many mathemeticians find them boring, and only ask those hot and unpredictable nonabelian groups out on dates.
mathwonk
#7
Feb29-12, 03:02 PM
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try this one:

Z/m x Z/n = Z/gcd x Z/lcm, whenever.


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