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Math Help - Group of order 30

  1. #1
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    Group of order 30

    I am working on showing that there are exactly 4 non-isomorphic groups of order 30. I'm stuck on showing that this group of order 30, G must have a normal subgroup, N of order 15.
    I have 30=2 \times 3 \times 5
    Then by using Sylow-theorems I have the possibilities for Sylow p-subgroups of G as followed:
    n_2=1,3,5,15;<br />
n_3=1,10;<br />
n_5=1,6
    Then I use counting argument to show that it can not be the case that n_5=6 and n_3=10. So, either n_5=1 or n_3=1. Then it follows that G has either a normal Sylow 3-subgroup or a normal Sylow 5-subgroup. But I can't argue that it must have both of those.
    I notice that by Cauchy, G must also have a subgroup of order 2, but didn't see how this helps.

    I really appreciate any help.
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  2. #2
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    Quote Originally Posted by jackie View Post
    I am working on showing that there are exactly 4 non-isomorphic groups of order 30. I'm stuck on showing that this group of order 30, G must have a normal subgroup, N of order 15.
    I have 30=2 \times 3 \times 5
    Then by using Sylow-theorems I have the possibilities for Sylow p-subgroups of G as followed:
    n_2=1,3,5,15;<br />
n_3=1,10;<br />
n_5=1,6
    Then I use counting argument to show that it can not be the case that n_5=6 and n_3=10. So, either n_5=1 or n_3=1. Then it follows that G has either a normal Sylow 3-subgroup or a normal Sylow 5-subgroup. But I can't argue that it must have both of those.
    I notice that by Cauchy, G must also have a subgroup of order 2, but didn't see how this helps.

    I really appreciate any help.
    let N_1,N_2 be two subgroups of G with |N_1|=3, \ |N_2|=5. you've proved that either N_1 or N_2 is normal. thus N_1N_2 is a subgroup of G. clearly |N_1N_2|=15 and so [G:N_1N_2]=2.

    now recall that every subgroup of index 2 in any group is normal.
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