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Is the direct sum of cyclic pgroups a cyclic group? 
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#1
Feb2912, 12:23 AM

P: 1,583

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. 


#2
Feb2912, 01:31 AM

Emeritus
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PF Gold
P: 4,500

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 p^{m}q^{n}?
Alternatively, this is an example of the chinese remainder theorem http://en.wikipedia.org/wiki/Chinese..._ideal_domains where R is the integers, u = p^{m}q^{n} 


#3
Feb2912, 02:36 AM

P: 1,583

Thanks Office_Shredder!



#4
Feb2912, 06:09 AM

Sci Advisor
P: 906

Is the direct sum of cyclic pgroups a cyclic group?
as a counterexample Z_{2} x Z_{2} 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 Z_{m} x Z_{n} cannot possibly be isomorphic to Z_{mn}, 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 Z_{mn} with Z_{m} x Z_{n} 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 preimage 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. 


#5
Feb2912, 09:34 AM

Mentor
P: 18,210

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 pgroups. 


#6
Feb2912, 09:58 AM

Sci Advisor
P: 906

abelian groups are the wellmannered groups that always say please and thankyou. as such, many mathemeticians find them boring, and only ask those hot and unpredictable nonabelian groups out on dates. 


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