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Failure of the sylow counting argument

  1. Apr 29, 2007 #1
    show that there is a normal subgroup of G of order 5 when G is a group of order 30. My friend just called me with this problem, he said the usual method of solution fails. (i.e. using sylow and then showing that the subgroup is unique and deducing that it must therfore be normal), I told him to post it on physics forum so ytou might get this question twice.
  2. jcsd
  3. Apr 29, 2007 #2


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    The theorem says that the number of 5-groups, n_5, divides 30/5=6 and is equal to 1 mod 5. So it's either 1 or 6.

    If it's 1, you're done, so assume it's 6. What would this imply about the number of elements of order 5 in the group? Try to show that it doesn't leave enough room for the other elements in the group.
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