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Thread: 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, $\displaystyle G$ must have a normal subgroup, $\displaystyle N$ of order 15.
    I have $\displaystyle 30=2 \times 3 \times 5$
    Then by using Sylow-theorems I have the possibilities for Sylow p-subgroups of G as followed:
    $\displaystyle n_2=1,3,5,15;
    n_3=1,10;
    n_5=1,6$
    Then I use counting argument to show that it can not be the case that $\displaystyle n_5=6$ and $\displaystyle n_3=10$. So, either $\displaystyle n_5=1$ or $\displaystyle n_3=1$. Then it follows that $\displaystyle G$ 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, $\displaystyle G$ must have a normal subgroup, $\displaystyle N$ of order 15.
    I have $\displaystyle 30=2 \times 3 \times 5$
    Then by using Sylow-theorems I have the possibilities for Sylow p-subgroups of G as followed:
    $\displaystyle n_2=1,3,5,15;
    n_3=1,10;
    n_5=1,6$
    Then I use counting argument to show that it can not be the case that $\displaystyle n_5=6$ and $\displaystyle n_3=10$. So, either $\displaystyle n_5=1$ or $\displaystyle n_3=1$. Then it follows that $\displaystyle G$ 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 $\displaystyle N_1,N_2$ be two subgroups of $\displaystyle G$ with $\displaystyle |N_1|=3, \ |N_2|=5.$ you've proved that either $\displaystyle N_1$ or $\displaystyle N_2$ is normal. thus $\displaystyle N_1N_2$ is a subgroup of $\displaystyle G.$ clearly $\displaystyle |N_1N_2|=15$ and so $\displaystyle [G:N_1N_2]=2.$

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