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Math Help - Minimal normal subgroups are elementary Abelian

  1. #1
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    Minimal normal subgroups are elementary Abelian

    Hey, can anybody help me with this?

    If G is a finite group with a minimal non-trivial normal subgroup M such that M is Abelian, then \exists p \in \mathbb{N}^+ prime such that \forall x \in M \backslash \{e\}, o(x) = p.

    I have no idea where to start with this. I can see that all proper non-trivial subgroups of M aren't normal in G, but not sure how this helps.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Giraffro View Post
    Hey, can anybody help me with this?

    If G is a finite group with a minimal non-trivial normal subgroup M such that M is Abelian, then \exists p \in \mathbb{N}^+ prime such that \forall x \in M \backslash \{e\}, o(x) = p.

    I have no idea where to start with this. I can see that all proper non-trivial subgroups of M aren't normal in G, but not sure how this helps.
    You have two things to prove, that only one prime divides the order of your subgroup and that your subgroup is a direct product of groups of order p.

    Hint: If H is a characteristic subgroup of K and K \lhd H then H \lhd G.
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  3. #3
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    Quote Originally Posted by Swlabr View Post
    You have two things to prove, that only one prime divides the order of your subgroup and that your subgroup is a direct product of groups of order p.

    Hint: If H is a characteristic subgroup of K and K \lhd H then H \lhd G.
    I haven't encountered characteristic subgroups before and this question is from a past paper a few years ago, so I may just skip it. Thanks for replying anyway.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Giraffro View Post
    I haven't encountered characteristic subgroups before and this question is from a past paper a few years ago, so I may just skip it. Thanks for replying anyway.
    I'm sure there are other ways, but that way seemed the neatest. Basically, a characteristic subgroup is a subgroup which is fixed under automorphisms,

    \phi : G \rigtharrow G then H\phi = H.

    Examples of characteristic subgroups are the two boring subgroups, the center, the commutator subgroup, and the frattini subgroup.

    If you take a cyclic group, then every subgroup is characteristic.

    Now, you can think of conjugation by an element as an automorphism, and so such groups are normal. However, the characteristic subgroup of a normal subgroup will be normal in the big group, and this is what I would have utilised.

    For the first part, if another prime divided the group M then take the largest q-group in M. As your group is abelian this will encompass all the q-ness of your group (q will not divide the order of M/Q). Therefore, you would have a characteristic subgroup, which will thus be normal.
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  5. #5
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    Quote Originally Posted by Swlabr View Post
    I'm sure there are other ways, but that way seemed the neatest. Basically, a characteristic subgroup is a subgroup which is fixed under automorphisms,

    \phi : G \rigtharrow G then H\phi = H.

    Examples of characteristic subgroups are the two boring subgroups, the center, the commutator subgroup, and the frattini subgroup.

    If you take a cyclic group, then every subgroup is characteristic.

    Now, you can think of conjugation by an element as an automorphism, and so such groups are normal. However, the characteristic subgroup of a normal subgroup will be normal in the big group, and this is what I would have utilised.

    For the first part, if another prime divided the group M then take the largest q-group in M. As your group is abelian this will encompass all the q-ness of your group (q will not divide the order of M/Q). Therefore, you would have a characteristic subgroup, which will thus be normal.
    So if I understand you correctly, this is how you prove it:

    Let p \in \mathbb{N}^+ be prime such that p \mid |M| and let P \in \text{Syl}_p(G) be given. Then since M is Abelian, P \lhd M and so |\text{Syl}_p(G)| = 1. Now let g \in G be given and define \phi : G \to G, h \mapsto h^g. Then \phi \in \text{Aut}(G) and since M \lhd G, \phi|_M \in \text{Aut}(M). Thus \phi(P) \in \text{Syl}_p(G) and so by uniqueness P = \phi(P) = P^g. Therefore P \lhd G and since M is a minimal non-trivial normal subgroup of G, P = M or P = \{e\}. However, p \mid |M|, so by Sylow's theorems, p \mid |P| and therefore P = M. Hence M is an Abelian p-group.

    Finally, define H := \{g \in M : o(g) \mid p\}. Then e \in H and as M is Abelian, \forall g, h \in M, o(gh) = lcm(o(g), o(h)) \mid p and o(g^{-1}) = o(g) \mid p, so gh, g^{-1} \in H. Therefore H \lhd M. Now let g \in G, h \in H be given. Then o(h^g)=o(h) \mid p, so since M \lhd G, h^g \in M and therefore h^g \in H. Thus H \lhd G and so H = M or H = \{e\}. However, by Cauchy's theorem \exists g \in M such that o(g) = p and thus g \in H \backslash \{e\}. Therefore M = H.

    Thanks for the help!
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Giraffro View Post
    So if I understand you correctly, this is how you prove it:

    Let p \in \mathbb{N}^+ be prime such that p \mid |M| and let P \in \text{Syl}_p(G) be given. Then since M is Abelian, P \lhd M and so |\text{Syl}_p(G)| = 1. Now let g \in G be given and define \phi : G \to G, h \mapsto h^g. Then \phi \in \text{Aut}(G) and since M \lhd G, \phi|_M \in \text{Aut}(M). Thus \phi(P) \in \text{Syl}_p(G) and so by uniqueness P = \phi(P) = P^g. Therefore P \lhd G and since M is a minimal non-trivial normal subgroup of G, P = M or P = \{e\}. However, p \mid |M|, so by Sylow's theorems, p \mid |P| and therefore P = M. Hence M is an Abelian p-group.
    Not quite...you are taking a sylow subgroup of M, not of G. There is no reason to say these coincide.

    Let p | |M|. As M is abelian, the sylow p-subgroups of M are normal and unique, so let P be the maximal subgroup of M of order a power of p. Clearly, if |M| = p^nq, p \not\div q then |P| = p^n.

    Now, we know that P \lhd M, but we wish to prove that P \lhd G. This is true because if \phi: M \rightarrow M is an automorphism, then P\phi = P because every element has order a power of p and no other element in the group has this order, and automorphisms preserve order (does that make sense)? Therefore, P \text{ char } M.

    Next, notice that for g \in G, ^g:M \rightarrow M, h \mapsto h^g is an automorphism of M. This is bacause M is normal in G. As this is an automorphism of M it must fix P. Therefore, P^g = P for all g \in G, and we thus conclude that P \lhd G, as required to get our contradiction.

    Note that K \lhd H \lhd G \not\Rightarrow K \lhd G.
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  7. #7
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    Quote Originally Posted by Swlabr View Post
    Not quite...you are taking a sylow subgroup of M, not of G. There is no reason to say these coincide.

    Let p | |M|. As M is abelian, the sylow p-subgroups of M are normal and unique, so let P be the maximal subgroup of M of order a power of p. Clearly, if |M| = p^nq, p \not\div q then |P| = p^n.

    Now, we know that P \lhd M, but we wish to prove that P \lhd G. This is true because if \phi: M \rightarrow M is an automorphism, then P\phi = P because every element has order a power of p and no other element in the group has this order, and automorphisms preserve order (does that make sense)? Therefore, P \text{ char } M.

    Next, notice that for g \in G, ^g:M \rightarrow M, h \mapsto h^g is an automorphism of M. This is bacause M is normal in G. As this is an automorphism of M it must fix P. Therefore, P^g = P for all g \in G, and we thus conclude that P \lhd G, as required to get our contradiction.

    Note that K \lhd H \lhd G \not\Rightarrow K \lhd G.
    I think there's a typo in my proof: I meant to say that \phi(P) \in \text{Syl}_p(M), since M \lhd G and as you say, automorphisms preserve group order.
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Giraffro View Post
    I think there's a typo in my proof: I meant to say that \phi(P) \in \text{Syl}_p(M), since M \lhd G and as you say, automorphisms preserve group order.
    Oh - it's clearly a typo too. I should have read your proof through properly. It would have saved about 15 minutes of typing...!
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  9. #9
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    Quote Originally Posted by Swlabr View Post
    Oh - it's clearly a typo too. I should have read your proof through properly. It would have saved about 15 minutes of typing...!
    No worries. Thanks for the help!
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