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    Is it true to say that if given two groups, with same number of elements and both groups have matching orders for all the elements, then they are isomorphic?

    So if group 1 has orders: 1,2,3,4,4,5
    and groups 2 also has orders: 1,2,3,4,4,5

    Does this imply they are isomorphic?

    I know that two isomorphic groups have the same order for all elements between the groups, but Im asking the converse.

    Thanks
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    (Original post by 2710)
    ...
    This may prove enlightening, but don't ask me any questions about it.
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    Well from that example, they seem to sy the answer is no.

    I think i foud my own exmple. D6 and C2xC3 have elements with the same order, but are not isomorphic.


    Am I right?
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    (Original post by 2710)
    Well from that example, they seem to sy the answer is no.

    I think i foud my own exmple. D6 and C2xC3 have elements with the same order, but are not isomorphic.


    Am I right?
    Don't think so.

    D6 has orders 1,2,2,2,3,3
    and c2xc3, which is isomorphic to c6 has orders 1,2,3,3,6,6

    Unless I made a slip.
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    This is a good question actually:

    No doubt, you can find a function that maps each element with the same order to each other, so if we have:

    G = \left\{a,b,c,d,e,p \right\}

    With respective orders 1,2,3,4,4,5 and

    H = \left\{h,i,j,k,l,m \right\}

    With the same respective orders 1,2,3,4,4,5

    Then you can find
    f_1: G \rightarrow H \ \ \ f_1(a) = h
    f_2: G \rightarrow H \ \ \ f_2(b) = i
    ...
    f_6: G \rightarrow H \ \ \ f_6(p) = m

    This just follows from the fact that any two cyclic groups of the same order are isomorphic.

    I'm wondering if you can define a function joining all these

    f : G \rightarrow H such that f(x) = \begin{cases}

            h, &         \text{if } x=a,\\

            i, &         \text{if } x = b.\\

j, &         \text{if } x=c.\\

k, &         \text{if } x=d.\\

l, &         \text{if } x=e.\\

m, &         \text{if } x=p.

    \end{cases}

    Which would be an isomorphism, but doing this seems somewhat abusive...
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    (Original post by Noble.)
    ...
    |G|=6 implies |g|\in\{1,2,3,6\}

    The rest made my brain hurt. Ouch!

    @OP: Missed that in your first post, but it wasn't really relevant to the thrust of your argument. But since it's reoccured....
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    (Original post by ghostwalker)
    |G|=6 implies |g|\in\{1,2,3,6\}

    The rest made my brain hurt. Ouch!
    I think I'm being dense here, how's this relevant?
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    (Original post by Noble.)
    I think I'm being dense here, how's this relevant?
    Not to the body of your argument, which I'm too tired to follow at present, but as it's repeating the OP's original erroneous orders, I thought it worth highlighting; for the benefit of others reading this thread and wondering what that group was, if nothing else. [/pedant]
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    (Original post by ghostwalker)
    Don't think so.

    D6 has orders 1,2,2,2,3,3
    and c2xc3, which is isomorphic to c6 has orders 1,2,3,3,6,6

    Unless I made a slip.
    You are right, I was thinking they both consisted of x^3 = 1 and y^2 = 1 so I jumped. but in fact, C is commutative, whereas D is not!

    Wait, so...i still haven't got the answer lol :P

    @Noble, so those two are isomorphic? So it is true?
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    (Original post by 2710)
    You are right, I was thinking they both consisted of x^3 = 1 and y^2 = 1 so I jumped. but in fact, C is commutative, whereas D is not!

    Wait, so...i still haven't got the answer lol :P

    @Noble, so those two are isomorphic? So it is true?
    Well, I seem to have found a bijective map which would indicate they are isomorphic. I'm sure someone will post later today either verifying my post or finding some major flaw in it.
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    (Original post by Noble.)
    Well, I seem to have found a bijective map which would indicate they are isomorphic. I'm sure someone will post later today either verifying my post or finding some major flaw in it.
    But in the link posted by ghostwalker, there are groups which have all the same orders (for their elements), yet are not isomorphic. So im confused, but i will wait dfor some more posts
 
 
 
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