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Math Help - Isomorphic Groups

  1. #1
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    Isomorphic Groups



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

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

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

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

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

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

    \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 \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?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    HINT: There is a theorem which says that \mathbb{Z}_{pq} \cong \mathbb{Z}_p \oplus \mathbb{Z}_q if and only if 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.
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  3. #3
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    Quote Originally Posted by demode View Post


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

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

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

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

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

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

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

    Tonio
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by tonio View Post
    Remember that \mathbb{Z}_{nm}\cong \mathbb{Z}_n \oplus \mathbb{Z}_m , whenever (n,m)=1 ...

    Tonio
    How is that at all different from what I said?
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  5. #5
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    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 \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2} has elements of order 1350, but \mathbb{Z}_{6} \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_{5} has no elements of order 1350. Thus

    \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?
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  6. #6
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    Quote Originally Posted by Swlabr View Post
    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...

    Tonio
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  7. #7
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    Quote Originally Posted by demode View Post
    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 \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2} has elements of order 1350, but \mathbb{Z}_{6} \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_{5} has no elements of order 1350. Thus

    \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

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

    Tonio
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  8. #8
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    I see you have broken \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2} down to \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 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.
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  9. #9
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by demode View Post
    I see you have broken \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2} down to \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 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 - \mathbb{Z}_{mn} \cong \mathbb{Z}_m \oplus \mathbb{Z}_n if and only if 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,

    \mathbb{Z}_{abc} \cong \mathbb{Z}_a \oplus \mathbb{Z}_b \oplus \mathbb{Z}_c if and only if 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,

    \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?
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  10. #10
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    Quote Originally Posted by Swlabr View Post
    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 - \mathbb{Z}_{mn} \cong \mathbb{Z}_m \oplus \mathbb{Z}_n if and only if 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,

    \mathbb{Z}_{abc} \cong \mathbb{Z}_a \oplus \mathbb{Z}_b \oplus \mathbb{Z}_c if and only if 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,

    \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 (a,b,c) \in \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2}, and since lcm(9,2,25)=1350, then (a,b,c) can have order 1350. However no element of such order exists in \mathbb{Z}_{6} \oplus \mathbb{Z}_{45} \oplus \mathbb{Z}_{5}. It's straightforward.
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  11. #11
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    Quote Originally Posted by demode View Post
    I see you have broken \mathbb{Z}_{3^3} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{5^2} down to \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 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.
    .
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