# Isomorphic Groups

• Sep 5th 2010, 09:07 PM
demode
Isomorphic Groups
http://img651.imageshack.us/img651/6149/34264939.gif

This is what I've done so far. For the first part of the problem, we decompose 1350. into a product of primes as $\displaystyle 3^3.2.5^2$. Then we have

$\displaystyle \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2}$

$\displaystyle \mathbb{Z}_{3^2} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2}$

$\displaystyle \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2}$

$\displaystyle \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}$

$\displaystyle \mathbb{Z}_{3^2} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}$

$\displaystyle \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}$

Is this right?

And what do I need to do for the second part? How do I identify the ones which are isomorphic to $\displaystyle \mathbb{Z}_{6} \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_{5}$?

In the last direct product in my list, gcd(3,3,3,2,5,5)=1 i.e. they are relatively prime. Does this mean it is isomorphic to the given group?
• Sep 5th 2010, 11:07 PM
Swlabr
HINT: There is a theorem which says that $\displaystyle \mathbb{Z}_{pq} \cong \mathbb{Z}_p \oplus \mathbb{Z}_q$ if and only if $\displaystyle gcd(p, q)=1$. If you haven't seen this before, it is probably a good idea to prove it, but it is definitely what you are expected to use here.
• Sep 6th 2010, 06:19 AM
tonio
Quote:

Originally Posted by demode
http://img651.imageshack.us/img651/6149/34264939.gif

This is what I've done so far. For the first part of the problem, we decompose 1350. into a product of primes as $\displaystyle 3^3.2.5^2$. Then we have

$\displaystyle \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2}$

$\displaystyle \mathbb{Z}_{3^2} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2}$

$\displaystyle \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2}$

$\displaystyle \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}$

$\displaystyle \mathbb{Z}_{3^2} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}$

$\displaystyle \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}$

Is this right?

Yes it is

And what do I need to do for the second part? How do I identify the ones which are isomorphic to $\displaystyle \mathbb{Z}_{6} \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_{5}$?

In the last direct product in my list, gcd(3,3,3,2,5,5)=1 i.e. they are relatively prime. Does this mean it is isomorphic to the given group?

Remember that $\displaystyle \mathbb{Z}_{nm}\cong \mathbb{Z}_n \oplus \mathbb{Z}_m$ , whenever $\displaystyle (n,m)=1$ ...

Tonio
• Sep 6th 2010, 06:23 AM
Swlabr
Quote:

Originally Posted by tonio
Remember that $\displaystyle \mathbb{Z}_{nm}\cong \mathbb{Z}_n \oplus \mathbb{Z}_m$ , whenever $\displaystyle (n,m)=1$ ...

Tonio

How is that at all different from what I said?
• Sep 7th 2010, 12:56 AM
demode
Yes, that is exactly the theorem we are expected to use! But could you please show me how to apply it in this situation? I have no idea how it can be applied here because both sides are direct products.

*I also thought of another approach: Since $\displaystyle \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2}$ has elements of order 1350, but $\displaystyle \mathbb{Z}_{6} \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_{5}$ has no elements of order 1350. Thus

$\displaystyle \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2} \ncong \mathbb{Z}_{6} \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_{5}$

Is this a correct reasoning?
• Sep 7th 2010, 02:26 AM
tonio
Quote:

Originally Posted by Swlabr
How is that at all different from what I said?

It is exactly what you said but, as many other times, somebody (I) didn't notice somebody else (you) already answered the question...(Wink)

Tonio
• Sep 7th 2010, 02:29 AM
tonio
Quote:

Originally Posted by demode
Yes, that is exactly the theorem we are expected to use! But could you please show me how to apply it in this situation? I have no idea how it can be applied here because both sides are direct products.

*I also thought of another approach: Since $\displaystyle \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2}$ has elements of order 1350, but $\displaystyle \mathbb{Z}_{6} \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_{5}$ has no elements of order 1350. Thus

$\displaystyle \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2} \ncong \mathbb{Z}_{6} \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_{5}$

Is this a correct reasoning?

It may be but it sure may also be a lengthy, suffering way. Note, for example, that

$\displaystyle \mathbb{Z}_6\cong \mathbb{Z}_3\oplus \mathbb{Z}_2\,,\,\,\mathbb{Z}_{45}\cong \mathbb{Z}_9\oplus\mathbb{Z}_5$ ...

Tonio
• Sep 8th 2010, 12:40 AM
demode
I see you have broken $\displaystyle \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2}$ down to $\displaystyle \mathbb{Z}_{3} \oplus \mathbb{Z}_{3^2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}$

You can do this because $\displaystyle gcd(27,25,2)=1$? And if they were not relatively prime you couldn't break it up like that? Right?

P.S. So your post shows that they are isomorphic? I'm confused, doesn't my previous post show that they are not isomorphic?

Thank you.
• Sep 8th 2010, 12:58 AM
Swlabr
Quote:

Originally Posted by demode
I see you have broken $\displaystyle \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2}$ down to $\displaystyle \mathbb{Z}_{3} \oplus \mathbb{Z}_{3^2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}$

You can do this because $\displaystyle gcd(27,25,2)=1$? And if they were not relatively prime you couldn't break it up like that? Right?

P.S. So your post shows that they are isomorphic? I'm confused, doesn't my previous post show that they are not isomorphic?

Thank you.

No, he hasn't broken down that group the way you said he did. Where did you get that from? What you said is NOT correct either - $\displaystyle \mathbb{Z}_{mn} \cong \mathbb{Z}_m \oplus \mathbb{Z}_n$ if and only if $\displaystyle gcd(m, n)=1$, but you cannot just assume this holds for products of length 3 or more - it will only hold in that form if they are pairwise coprime,

$\displaystyle \mathbb{Z}_{abc} \cong \mathbb{Z}_a \oplus \mathbb{Z}_b \oplus \mathbb{Z}_c$ if and only if $\displaystyle gcd(a, b) = gcd(b, c) = gcd(a, c) = 1$.

The groups you want are NOT isomorphic. The way you did it is the long way. Although correct, it is much easier just to use the two equalities he gave you. This will give you,

$\displaystyle \mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 \cong \mathbb{Z}_2 \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_9 \oplus \mathbb{Z}_5\oplus \mathbb{Z}_{5}$.

Can you finish from here?
• Sep 12th 2010, 02:55 AM
demode
Quote:

Originally Posted by Swlabr
No, he hasn't broken down that group the way you said he did. Where did you get that from? What you said is NOT correct either - $\displaystyle \mathbb{Z}_{mn} \cong \mathbb{Z}_m \oplus \mathbb{Z}_n$ if and only if $\displaystyle gcd(m, n)=1$, but you cannot just assume this holds for products of length 3 or more - it will only hold in that form if they are pairwise coprime,

$\displaystyle \mathbb{Z}_{abc} \cong \mathbb{Z}_a \oplus \mathbb{Z}_b \oplus \mathbb{Z}_c$ if and only if $\displaystyle gcd(a, b) = gcd(b, c) = gcd(a, c) = 1$.

The groups you want are NOT isomorphic. The way you did it is the long way. Although correct, it is much easier just to use the two equalities he gave you. This will give you,

$\displaystyle \mathbb{Z}_6 \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_5 \cong \mathbb{Z}_2 \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_9 \oplus \mathbb{Z}_5\oplus \mathbb{Z}_{5}$.

Can you finish from here?

Thanks. So they are not isomorphic because they are not pairwise coprime since 3 and 9 are not coprime, also 5 and 5 are not coprime. Is this right?

P.S. Why did you say my method is the long way? All I did was note that if $\displaystyle (a,b,c) \in \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2}$, and since $\displaystyle lcm(9,2,25)=1350$, then (a,b,c) can have order 1350. However no element of such order exists in $\displaystyle \mathbb{Z}_{6} \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_{5}$. It's straightforward.
• Sep 12th 2010, 03:49 AM
tonio
Quote:

Originally Posted by demode
I see you have broken $\displaystyle \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2}$ down to $\displaystyle \mathbb{Z}_{3} \oplus \mathbb{Z}_{3^2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{5}$

No, I haven't...why do you think I have?? Please do read carefully what I, and YOU, wrote.

Tonio

You can do this because $\displaystyle gcd(27,25,2)=1$? And if they were not relatively prime you couldn't break it up like that? Right?

P.S. So your post shows that they are isomorphic? I'm confused, doesn't my previous post show that they are not isomorphic?

Thank you.

.