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Homework Help: Groups of order 30

  1. Jan 19, 2009 #1
    Is it true that any group of order 30 has a normal (hence unique) Sylow-5 subgroup?

    I know that that the only possibilities for n(5) are 1 or 6.

    Now suppose there are 6 sylow 5 subgroups in G. This would yield
    (5-1)6=24 distinct elements of order 5 in G. Now there is only 30-24=6
    elements left in G and one of these is the identity. This means that there
    must be 1 sylow 3 subgroup in G which has 2 distinct elements of order 2
    now there is only 4 elements left in G one of them being the identity
    so there must be 3 sylow 2 subgroups of G in this case each having 1 distinct
    element of order 2

    I don't see where is the contradiction here.

    The only thing I know is that there must be a cyclic normal subgroup of order 15 in G.
    Last edited: Jan 19, 2009
  2. jcsd
  3. Jan 19, 2009 #2
    Oh I think I got it.

    Every group G of order 30 has a normal cyclic subgroup of order 15 (I can prove this).
    Let's call it H.
    Now, consider Syl_5(H).
    n_5=1, hence if P lies in Syl_5(H), then P is the unique normal subgroup of H of order 5. Hence P char H. Since H is normal in G, it follows P is normal in G.

    Now |P|=5 and |G|=(2^2)*3*5. So P is a normal Sylow-5 subgroup of G.
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