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Math Help - Isomorphisms

  1. #1
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    Isomorphisms

    Is there any simple way of telling if two groups are isomorphic? For example, given a list like this and asked to partition into isomorphism classes

    1. \mathbb{Z}_{4} \times \mathbb{Z}_{2}
    2. \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}
    3. Q_{8}
    4. D_{8}
    5. \mathbb{Z}_{8}
    6. \powerset \left\{ 1,2,3 \right\} under symmetric set difference \left( X \cup Y \right) \backslash \left( X \cap Y \right)
    7. Solutions of x^{8}-1=0 in  \mathbb{C}
    8. \left< \left\( 24\right\), \left( 12 \right) \left( 34 \right)   \right> \leq S_{4}
    9. \left< \left[ {\begin{array}{cc} i & 0  \\ 0 & -i  \\ \end{array} } \right], \left[ {\begin{array}{cc} 0 & i  \\ i & 0  \\ \end{array} } \right], \left[ {\begin{array}{cc} 0 & -1  \\ 1 & 0  \\ \end{array} } \right] \right> \subseteq \mathbb{C}^{2\times2}
    10.  \mathbb{Z}_{16}^{*}

    Can I work out if they are isomorphic without listing out the elements of each, trying to find a map to from each one to another etc?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Conn View Post
    Is there any simple way of telling if two groups are isomorphic? For example, given a list like this and asked to partition into isomorphism classes

    1. \mathbb{Z}_{4} \times \mathbb{Z}_{2}
    2. \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}
    3. Q_{8}
    4. D_{8}
    5. \mathbb{Z}_{8}
    6. \powerset \left\{ 1,2,3 \right\} under symmetric set difference \left( X \cup Y \right) \backslash \left( X \cap Y \right)
    7. Solutions of x^{8}-1=0 in  \mathbb{C}
    8. \left< \left\( 24\right\), \left( 12 \right) \left( 34 \right)   \right> \leq S_{4}
    9. \left< \left[ {\begin{array}{cc} i & 0  \\ 0 & -i  \\ \end{array} } \right], \left[ {\begin{array}{cc} 0 & i  \\ i & 0  \\ \end{array} } \right], \left[ {\begin{array}{cc} 0 & -1  \\ 1 & 0  \\ \end{array} } \right] \right> \subseteq \mathbb{C}^{2\times2}
    10.  \mathbb{Z}_{16}^{*}

    Can I work out if they are isomorphic without listing out the elements of each, trying to find a map to from each one to another etc?
    I mean there is no simple way in general. So, for example \mathbb{Z}_4\times\mathbb{Z}_2\not\cong\mathbb{Z}_  2^3 since the former has the element (1,0) of order four and the latter has no element of order 4. It's easy that \mathbb{Z}_4\times\mathbb{Z}_2\not\cong Q_8,D_8, since the latter two aren't abelian. The fact that D_8\not\cong Q_8 can be gotten from counting element orders.


    One can easily see that \left\{x\in\mathbb{C}:x^8=1\right\}\cong\mathbb{Z}  _8 (since it's cyclic).


    You can check that 2^{\{1,2,3\}} with symmetric difference is isomorphic to \mathbb{Z}_2^3 since it's abelian and A^2=A\Delta A=\varnothing for every A\in 2^{\{1,2,3\}}



    Try a few more and see if you can work it out. Things to check are number of subgroups of a given order, number of normal subgroups of a given order, number of elements of a given order, whether the two groups are abelian, whether they are cyclic, etc. are all properties preserved under isomorphism.
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  3. #3
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    Quote Originally Posted by Conn View Post
    Is there any simple way of telling if two groups are isomorphic? For example, given a list like this and asked to partition into isomorphism classes

    1. \mathbb{Z}_{4} \times \mathbb{Z}_{2}
    2. \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}
    3. Q_{8}
    4. D_{8}
    5. \mathbb{Z}_{8}
    6. \powerset \left\{ 1,2,3 \right\} under symmetric set difference \left( X \cup Y \right) \backslash \left( X \cap Y \right)
    7. Solutions of x^{8}-1=0 in  \mathbb{C}
    8. \left< \left\( 24\right\), \left( 12 \right) \left( 34 \right)   \right> \leq S_{4}
    9. \left< \left[ {\begin{array}{cc} i & 0  \\ 0 & -i  \\ \end{array} } \right], \left[ {\begin{array}{cc} 0 & i  \\ i & 0  \\ \end{array} } \right], \left[ {\begin{array}{cc} 0 & -1  \\ 1 & 0  \\ \end{array} } \right] \right> \subseteq \mathbb{C}^{2\times2}
    10.  \mathbb{Z}_{16}^{*}

    Can I work out if they are isomorphic without listing out the elements of each, trying to find a map to from each one to another etc?

    I think the easiest thing to do here is to sort out groups that can't be isomorphic. For example, (1) and (2) can't be iso.

    since one has elements of order 4 whereas the second one hasn't, and none is iso. with (3) since this last one isn't

    abelian whereas the first two are.

    Continue in this fashion until you have fewer cases to check.

    Tonio
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  4. #4
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    basically, what you have to do create finer and finer equaivalences on your 10-element set. one of the coarsest filters, if you will, is to sort finite groups by order. here, all 10 groups are of order 8, so that isn't particularly helpful.

    the next partition would be abelian versus non-abelian. this gives us 2 subsets {1,2,5,6,7,10} and {3,4,8,9}.

    for the abelian groups, we can subdivide by cyclic versus non-cyclic: {1,2,6,10} and {5,7}. since cyclic groups of the same order are isomorphic, {5,7} is one equivalence class under isomorphism.

    by considering orders of elements, we can further sub-dvide {1,2,6,10} into {1,10} and {2,6}. Drexel28 outlined how you can define an isomorphism between
    P({1,2,3}) and Z2 x Z2 x Z2, so {2,6} is another equivalence class. in fact U(Z16) is isomorphic to Z4 x Z2, so {1,10} is a 3rd equivalence class (to actually prove this, you might try to exhibit an isomorphism. i suggest finding an element x of order 4 in U(Z16) and an order y of order 2 in U(Z16), and sending
    (a,b) --> (x^a)(y^b) mod 16).

    so that leaves us with {3,4,8,9}. considering orders of elements shows that 3 and 9 have only 1 element of order 2, while 4 and 8 have more than 1. finding an explicit isomorphism between D4 and <(2 4), (1 2)(3 4)> shouldn't be hard (some effort has been made to disguise the nature of the latter group. (1 2)(3 4)(2 4) = (1 2 3 4), which is a 4-cycle).

    by a suitable assignment of i,j,k to the 3 2x2 complex matrices shown, you should be able to show the last isomorphism equivalence class is {3,9}.
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