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jango_IN_DTOWN

  • one year ago

Subgroup + permutation problem

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  1. jango_IN_DTOWN
    • one year ago
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  2. jango_IN_DTOWN
    • one year ago
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    @ganeshie8

  3. ganeshie8
    • one year ago
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    Hey still here ?

  4. jango_IN_DTOWN
    • one year ago
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    hii

  5. ganeshie8
    • one year ago
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    familiar with the cycle notation ? |dw:1444581237571:dw|

  6. jango_IN_DTOWN
    • one year ago
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    yes |dw:1444581322888:dw|

  7. ganeshie8
    • one year ago
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    Ah, so is it given that \((1,2)\in S_3\) ?

  8. ganeshie8
    • one year ago
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    |dw:1444581438396:dw|

  9. jango_IN_DTOWN
    • one year ago
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    no i assumed it to be S3 since the next two contains 3, we are not given about the symmetry group

  10. ganeshie8
    • one year ago
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    since \(\alpha\) exists in the subgroup, \(\alpha^2, \alpha^3, \ldots \) must also exist in the group. \(\alpha^2 = ?\)

  11. jango_IN_DTOWN
    • one year ago
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    |dw:1444581538927:dw|

  12. jango_IN_DTOWN
    • one year ago
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    the identity permutation

  13. ganeshie8
    • one year ago
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    so we have \(\alpha^2 = e\), therefore the subgroup generated by \((1,2)\) is : \(\{e,\alpha\}\)

  14. jango_IN_DTOWN
    • one year ago
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    hmm true...

  15. ganeshie8
    • one year ago
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    you may express the same in cycle notation as \(\{(1), (1,2)\}\)

  16. ganeshie8
    • one year ago
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    try working ii similarly, just consider all the powers of given permutation

  17. jango_IN_DTOWN
    • one year ago
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    ok thanks...

  18. jango_IN_DTOWN
    • one year ago
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    we will stop when the identity permutation is reached

  19. ganeshie8
    • one year ago
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    elements will be repeated if you continue, notice that \(\alpha^n =e \implies \alpha^{n+1}=\alpha \)

  20. jango_IN_DTOWN
    • one year ago
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    Yes I got it.. Thanks..

  21. ganeshie8
    • one year ago
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    np

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