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Math Help - group isomorphism

  1. #1
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    group isomorphism

    If G is a group and n>= 1, define G(n) = {x in G : ord(x) = n}

    If G =~ H ( isomorphism ), show that for all n>= 1, |G(n)| = |H(n)|

    deduce that C3* C3 is not isomorphic C9 ( where C3, C9 are the cyclic groups of order 3 and 9 respectively)

    I have no clue how to do this question. Please help

    Is it true that C3 * C5 =~ C15? I think the answer is YES because 3 and 5 are prime numbers, so the groups of these order are cyclic. Am I right?

    Is it true that C2 * C6 =~ C12? I think not maybe because 6 is not a prime number.

    Thank you for your time, I'm really appreciated
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  2. #2
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    Quote Originally Posted by knguyen2005 View Post
    If G is a group and n>= 1, define G(n) = {x in G : ord(x) = n}

    If G =~ H ( isomorphism ), show that for all n>= 1, |G(n)| = |H(n)|

    deduce that C3* C3 is not isomorphic C9 ( where C3, C9 are the cyclic groups of order 3 and 9 respectively)
    If \phi :G\to H is an isomorphism then the order of x is same as order of \phi(x). Thus, define a mapping from G(n) to H(n) by a\mapsto \phi(a) to show |G(n)| = |H(n)|.

    Now to show C_3\times C_3\not \simeq C_9 show that |C_3\times C_3(3) |\not = |C_9 (3)|.
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  3. #3
    Junior Member ursa's Avatar
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    If G is a group and n>= 1, define G(n) = {x in G : ord(x) = n}

    If G =~ H ( isomorphism ), show that for all n>= 1, |G(n)| = |H(n)|

    deduce that C3* C3 is not isomorphic C9 ( where C3, C9 are the cyclic groups of order 3 and 9 respectively)

    I have no clue how to do this question. Please help

    Is it true that C3 * C5 =~ C15? I think the answer is YES because 3 and 5 are prime numbers, so the groups of these order are cyclic. Am I right?

    Is it true that C2 * C6 =~ C12? I think not maybe because 6 is not a prime number.

    Thank you for your time, I'm really appreciated
    basically its a property that if G =~ H ( isomorphism ) then |G(n)| = |H(n)|
    if you need a proof then do tell i wlll post it
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  4. #4
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    To Ursa

    Quote Originally Posted by ursa View Post
    basically its a property that if G =~ H ( isomorphism ) then |G(n)| = |H(n)|
    if you need a proof then do tell i wlll post it
    To Ursa, can you post me a proof of this problem please?
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  5. #5
    o_O
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    Well, if two groups in general are isomorphic then by definition there exists a bijective mapping from one group to the other.

    In order for us to have this mapping, we must have that both groups have the same number of elements (two sets have the same number of elements iff there exists a bijection between them).

    Thus, the two groups must have the same order.
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  6. #6
    Lord of certain Rings
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    Quote Originally Posted by knguyen2005 View Post
    To Ursa, can you post me a proof of this problem please?
    I wonder why you ask when, ThePerfectHacker has clearly explained in post #2 that:

    Quote Originally Posted by ThePerfectHacker View Post
    If \phi :G\to H is an isomorphism then the order of x is same as order of \phi(x). Thus, define a mapping from G(n) to H(n) by a\mapsto \phi(a) to show |G(n)| = |H(n)|.

    Now to show C_3\times C_3\not \simeq C_9 show that |C_3\times C_3(3) |\not = |C_9 (3)|.
    Is it true that C2 * C6 =~ C12? I think not maybe because 6 is not a prime number.
    All elements x \in C_2 \times C_6 have the property that x^6 = 1 (Verify!). But C_{12} needs has at least one element with order 12. Thus C_{12} \not \cong C_2 \times C_6
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  7. #7
    Junior Member ursa's Avatar
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    To Ursa, can you post me a proof of this problem please?

    hi
    well yes ThePerectHacker and o_O has clearly explained

    i m jst converting in mathematical form

    X Є G
    and O(G)=n
    => X^n =e
    => f(X^n)=f(e)
    => f(X.X.X.X.X….X)=e’
    => f(X). f(X) . f(X) . f(X) . f(X)…… . f(X)=e’
    => (f(X))^n=e’
    => O(f(X))=n
    =>O(H)=n
    =>O(H)=O(G)
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  8. #8
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    Since no one has answered the latter questions I will give you a hint. Look at  C_3 \times C_5 Your intuition is right that this is isomorphic to C_{15}. But why? Your hunch was because 3 and 5 were prime. But notice that C_3 \times C_3 \not \simeq C_9 because C_3 \times C_3 is not cyclic.

    But what else do we notice about 3 and 5?
    Last edited by mr fantastic; January 17th 2009 at 09:39 PM. Reason: Fixed the latex. To the OP: eg. Spacing is needed between distinct latex commands such as \times
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  9. #9
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    I think its important to point out whats going on behind the scenes of gropus of the form C_m\times{C_n}. Maybe proving or disproving this claim can give you some insight.

    Claim: C_m\times{C_n}\simeq{C_{mn}} if and only if  gcd(m,n)=1
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