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Math Help - Groups, subgroups and indexes

  1. #1
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    Groups, subgroups and indexes

    Here is a problem I think I have a solution for, I am just not 100% sure on one of the detail

    Let G be a group that does not have a subgroup of index 2.
    Show that every subgroup of index 3 is normal in G

    My proof

    Let H\leq G, [G:H]=3, first we will assume that H is not trivial (if it is then it is normal).
    By cayley's expanded theorem, there exists a homomorphism from G to the permutation group on S ( S is the right coset of H). So this homomorphism h is from G to S_{3}. now we will look at K=kerh, according to cayley's expanded theorem K is normal in G, but since according to cayley's theorem every group is isomorphic to a subgroup of S_{n}, then we can say that K\lhd S_{n}. Now since we assumed H is not trivial and [G:H]=3 we get that n>5. But the only normal subgroup of S_{n} in this case is A_{n}, then we get [G:K]=2 and thats a contradiction.

    The only part I am not sure about is where I use S_{n} and G interchangeably, therfore saying that if K\lhd G then K=A_{n}. Can I use this reasoning?

    Appreciate your opinions.

    SK
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by skyking View Post
    Here is a problem I think I have a solution for, I am just not 100% sure on one of the detail

    Let G be a group that does not have a subgroup of index 2.
    Show that every subgroup of index 3 is normal in G

    My proof

    Let H\leq G, [G:H]=3, first we will assume that H is not trivial (if it is then it is normal).
    By cayley's expanded theorem, there exists a homomorphism from G to the permutation group on S ( S is the right coset of H). So this homomorphism h is from G to S_{3}. now we will look at K=kerh, according to cayley's expanded theorem K is normal in G, but since according to cayley's theorem every group is isomorphic to a subgroup of S_{n}, then we can say that K\lhd S_{n}. Now since we assumed H is not trivial and [G:H]=3 we get that n>5. But the only normal subgroup of S_{n} in this case is A_{n}, then we get [G:K]=2 and thats a contradiction.

    The only part I am not sure about is where I use S_{n} and G interchangeably, therfore saying that if K\lhd G then K=A_{n}. Can I use this reasoning?

    Appreciate your opinions.

    SK
    What seems off is that you claim that K\unlhd S_n. What does that even mean? Isn't K\subseteq G Oh, you just mean to identify it with it's image, you should state that explicitly. There's still an issue since in general the image of a normal subgroup under a non-epimorphism needn't be normal.
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  3. #3
    Senior Member roninpro's Avatar
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    I would start from the point before Drexel's objection and examine \text{ker}(h). You should try to determine what is in it.
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  4. #4
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    Quote Originally Posted by skyking View Post
    Here is a problem I think I have a solution for, I am just not 100% sure on one of the detail

    Let G be a group that does not have a subgroup of index 2.
    Show that every subgroup of index 3 is normal in G

    My proof

    Let H\leq G, [G:H]=3, first we will assume that H is not trivial (if it is then it is normal).
    By cayley's expanded theorem, there exists a homomorphism from G to the permutation group on S ( S is the right coset of H). So this homomorphism h is from G to S_{3}. now we will look at K=kerh, according to cayley's expanded theorem K is normal in G, but since according to cayley's theorem every group is isomorphic to a subgroup of S_{n}, then we can say that K\lhd S_{n}. Now since we assumed H is not trivial and [G:H]=3 we get that n>5. But the only normal subgroup of S_{n} in this case is A_{n}, then we get [G:K]=2 and thats a contradiction.

    The only part I am not sure about is where I use S_{n} and G interchangeably, therfore saying that if K\lhd G then K=A_{n}. Can I use this reasoning?

    Appreciate your opinions.

    SK
    Based on your notation, let S be the set of all left cosets of H in G. Now let G act on S by left translation. Let h be the induced homomorphism by this action h:G \rightarrow A(S) given by g \mapsto \tau_g (see Hungerford p91), and let K be the kernel of h. Since [G:H]=3 by hypothesis, we see that A(S) is S_3.

    The key argument should be G/K is isomorphic to a subgroup of S_3. Thus, |G/K|=[G:K] must divide |S_3|=6 by Lagrange's theorem. The possible choice of |G/K| is 1,2,3, and 6. Since G has no subgroup of index 2 by hypothesis and K \leq H (verify this), the only possible choice of |G/K| is 3 (verify this). Since [G:K]=[G:H][H:K], we conclude that H=K. Thus, H is normal in G.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    What seems off is that you claim that K\unlhd S_n. What does that even mean? Isn't K\subseteq G Oh, you just mean to identify it with it's image, you should state that explicitly. There's still an issue since in general the image of a normal subgroup under a non-epimorphism needn't be normal.
    Drexel,thanks for the reply.

    OK, can I "fix" it by saying then that since G\cong S_{n} and A_{n} is the only normal subgroup of S_{n} then it must hold that K\cong A_{n} so therfore they have the same order and the same index in their respective groups?
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by skyking View Post
    Drexel,thanks for the reply.

    OK, can I "fix" it by saying then that since G\cong S_{n} and A_{n} is the only normal subgroup of S_{n} then it must hold that K\cong A_{n} so therfore they have the same order and the same index in their respective groups?
    ...but G \not\cong S_n...
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  7. #7
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    Quote Originally Posted by skyking View Post
    Drexel,thanks for the reply.

    OK, can I "fix" it by saying then that since G\cong S_{n} and A_{n} is the only normal subgroup of S_{n} then it must hold that K\cong A_{n} so therfore they have the same order and the same index in their respective groups?
    No because, as has already been said, G\ncong S_n, and K is inside of H, NOT inside of S_n. Also, identifying K with its image in S_n wouldn't be too useful since, by definition, its image is \{0\}.

    Read the post above the one that I quoted, it has the right ideas.
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