# Direct products and isomorphisms

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• Apr 3rd 2013, 05:09 PM
Salome
Direct products and isomorphisms
By the Chinese Remainder Theorem,

if gcd(p,q) > 1, then ℤp x ℤq ≇ ℤpq

Does this imply that if ℤa x ℤb ≇ ℤc and ℤux ℤv ≇ ℤc, then ℤa x ℤb ≇ ℤux ℤv ?

Or is there something I'm missing?

Thanks!
• Apr 3rd 2013, 05:42 PM
Nehushtan
Re: Direct products and isomorphisms
In general, $\mathbb Z_m\times\mathbb Z_n\cong\mathbb Z_{\mathrm{lcm}(m,n)}$.
• Apr 3rd 2013, 06:01 PM
Salome
Re: Direct products and isomorphisms
Yes, but what I wanted to check was if there are two different direct products both not isomorphic to the same ℤ group, does that imply that the two different direct products are also not isomorphic to each other?
• Apr 3rd 2013, 07:40 PM
johng
Re: Direct products and isomorphisms
Hi,
First, I think Nehushtan's response is a little misleading. If m and n are not relatively prime, the direct product $\mathbb Z_m\times \mathbb Z_n$ is not cyclic.
Next, I really don't know what you're asking. The relation "not isomorphic" is never transitive.
• Apr 3rd 2013, 07:55 PM
Salome
Re: Direct products and isomorphisms
I've been delving into the theory of it and I think I'm starting to understand a bit more, I probably didn't phrase my original question very well. I'm trying to work out if a direct product of two ℤ groups is isomorphic to the direct product of two different ℤ groups and I was wondering if it could be generalised. In this case, the order of both direct products is the same, but the gcd and lcm of each of them are different.

My thought at this point is that they aren't isomorphic because the orders of their elements won't be the same if they have different gcd and lcm.
• Apr 5th 2013, 05:49 PM
Nehushtan
Re: Direct products and isomorphisms
Quote:

Originally Posted by johng
I think Nehushtan's response is a little misleading. If m and n are not relatively prime, the direct product $\mathbb Z_m\times \mathbb Z_n$ is not cyclic.

Doh! Of course. I must have confused the cyclic groups $\mathbb Z_m,\mathbb Z_n$ with the subgroups $m\mathbb Z,n\mathbb Z$ of $\mathbb Z$. (Doh)

Sorry for the misleading information. (Itwasntme)