jango_IN_DTOWN
  • jango_IN_DTOWN
Subgroup + permutation problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jango_IN_DTOWN
  • jango_IN_DTOWN
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jango_IN_DTOWN
  • jango_IN_DTOWN
@ganeshie8
ganeshie8
  • ganeshie8
Hey still here ?

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jango_IN_DTOWN
  • jango_IN_DTOWN
hii
ganeshie8
  • ganeshie8
familiar with the cycle notation ? |dw:1444581237571:dw|
jango_IN_DTOWN
  • jango_IN_DTOWN
yes |dw:1444581322888:dw|
ganeshie8
  • ganeshie8
Ah, so is it given that \((1,2)\in S_3\) ?
ganeshie8
  • ganeshie8
|dw:1444581438396:dw|
jango_IN_DTOWN
  • jango_IN_DTOWN
no i assumed it to be S3 since the next two contains 3, we are not given about the symmetry group
ganeshie8
  • ganeshie8
since \(\alpha\) exists in the subgroup, \(\alpha^2, \alpha^3, \ldots \) must also exist in the group. \(\alpha^2 = ?\)
jango_IN_DTOWN
  • jango_IN_DTOWN
|dw:1444581538927:dw|
jango_IN_DTOWN
  • jango_IN_DTOWN
the identity permutation
ganeshie8
  • ganeshie8
so we have \(\alpha^2 = e\), therefore the subgroup generated by \((1,2)\) is : \(\{e,\alpha\}\)
jango_IN_DTOWN
  • jango_IN_DTOWN
hmm true...
ganeshie8
  • ganeshie8
you may express the same in cycle notation as \(\{(1), (1,2)\}\)
ganeshie8
  • ganeshie8
try working ii similarly, just consider all the powers of given permutation
jango_IN_DTOWN
  • jango_IN_DTOWN
ok thanks...
jango_IN_DTOWN
  • jango_IN_DTOWN
we will stop when the identity permutation is reached
ganeshie8
  • ganeshie8
elements will be repeated if you continue, notice that \(\alpha^n =e \implies \alpha^{n+1}=\alpha \)
jango_IN_DTOWN
  • jango_IN_DTOWN
Yes I got it.. Thanks..
ganeshie8
  • ganeshie8
np

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