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Math Help - Group of order 21 - prove it's not comutative but solvable

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    Group of order 21 - prove it's not comutative but solvable

    Let G be a group of order 21, with at least 3 elements of order 3. Prove that G is not commutative, but solvable.

    I believe it's got something to do with the 3-Sylow sub-group of G, and there can be either 1 3-Sylow-subgroup or 7 3-Sylow-subgroup. How do I continue this ?

    Thanks!
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    Quote Originally Posted by adam63 View Post
    Let G be a group of order 21, with at least 3 elements of order 3. Prove that G is not commutative, but solvable.

    I believe it's got something to do with the 3-Sylow sub-group of G, and there can be either 1 3-Sylow-subgroup or 7 3-Sylow-subgroup. How do I continue this ?

    Thanks!

    A commutative group of order 21=3\cdot 7 has normal Sylow subgroups of order 7 and 3 and thus one of

    each, from where it follows that if our group has three different groups of order 3 it can't be commutative.


    Nevertheless, such a group does have one single Sylow 7-subgroup P which is then normal, and then

    we get that both P\,\,\,and\,\,\,G/P\cong C_3 are solvable groups (even abelian), and thus also G is.

    Tonio
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    hmm.. but if n_3 divides 7=21/3, and n_3=1mod(7), then n_3 is either 1 or 7. Why do you claim it has three different groups of order 3? How comes it doesn't have only one?

    And, how do you use the fact that there are at least 3 elements of order 3 in the group, in order to find out there exist three different groups of order 3?

    Thanks
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    Quote Originally Posted by adam63 View Post
    hmm.. but if n_3 divides 7=21/3, and n_3=1mod(7), then n_3 is either 1 or 7. Why do you claim it has three different groups of order 3? How comes it doesn't have only one?


    My bad: i misread the question. You said there're three different elements of order 3 and I read "three different subgroups of order 3". It never minds: if there are more than 2 elements of order 3 then there are more than 1 subgroup of order 3 and thus the argument stays the same.
    And you're right: there are either 1 or 7 subgroups of order 3 in a group of order 21...

    Tonio


    And, how do you use the fact that there are at least 3 elements of order 3 in the group, in order to find out there exist three different groups of order 3?

    Thanks
    .
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    I still can't seem to understand how that gets me closer to the solution - my question stays the same
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    Quote Originally Posted by adam63 View Post
    I still can't seem to understand how that gets me closer to the solution - my question stays the same

    It's written there: if a group of order 21 has more than 2 elements of order 3 then it has more than 1 unique (Sylow) 3-subgroup and thus it can't be abelian.

    Since any group of order 21 has one unique (Sylow) 7-sbgp. P this sbgp. is normal, and since P and G/P
    are

    solvable then G is solvable, or even easier: look at the series 1\leq P \leq G ...

    Tonio
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