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Math Help - Need help with a proof - subgroups

  1. #1
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    Need help with a proof - subgroups

    Hi guys.
    I need to prove the following statment, I did most of the work (which I'm not entirely sure is valid - so a feedback would be great), but I'm having some trouble proving the last part of it, and could really use some tips here.

    Let G be a group.
    Let H,K be non-trivial subgroups of G ( \left \{ e \right \} <H,K<G).
    Prove that H\cup K\neq G.

    Proof:
    If H=K we're done, since H\cup K=H, and H<G, we get that there exist g\in G s.t g\notin H\cup K.

    If H\neq K, without loss of generality, suppose that K<H.

    Let h\neq e\in H s.t h\notin K (there exist such h since H isn't trivial and K<H).

    Let k\in K.

    Obviusly h\cdot k\in G, so If we show that h\cdot k\notin H\cup K, we're done.

    Let's assume by contradiction that h\cdot k\in H\cup K.

    It then follows that h\cdot k\in H \vee h\cdot k\in K.

    Case I ( h\cdot k\in K):

    If h\cdot k\in K, then h\cdot k=m, for some m\in K.

    But then:

    h\cdot k=m

    h\cdot k\cdot k^{-1}=m\cdot k^{-1}

    h=m\cdot k^{-1}\newline

    but m\cdot k^{-1}\in K \Rightarrow h\in K! that contradics our assumption of h\notin K.

    Case II ( h\cdot k\in H):
    I have nothing here...


    any help would be greatly appreciated!
    Last edited by Stormey; November 19th 2013 at 12:02 PM.
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    Re: Need help with a proof - subgroups

    Quote Originally Posted by Stormey View Post
    Hi guys.
    I need to prove the following statment, I did most of the work (which I'm not entirely sure is valid - so a feedback would be great), but I'm having some trouble proving the last part of it, and could really use some tips here.

    Let G be a group.
    Let H,K be non-trivial subgroups of G ( \left \{ e \right \} <H,K<G).
    Prove that H\cup K\neq G.

    Proof:
    If H=K we're done, since H\cup K=H, and H<G, we get that there exist g\in G s.t g\notin H\cup K.

    If H\neq K, without loss of generality, suppose that K<H.
    I see reason that K would have to be a subgroup of H. Consider the symmetries of a square, D_8. H = {e, reflection over the x - axis} and K = {e, reflection over the y - axis} are both subgroups of D_8, but K is not a subgroup of H.

    -Dan
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    Re: Need help with a proof - subgroups

    Quote Originally Posted by topsquark View Post
    I see reason that K would have to be a subgroup of H. Consider the symmetries of a square, D_8. H = {e, reflection over the x - axis} and K = {e, reflection over the y - axis} are both subgroups of D_8, but K is not a subgroup of H.

    -Dan
    Hi, topsquark, thanks for the help.
    Yes of course, K does not have to be a subgroup of H.
    but there has to be an element in H that doesn't appear in K (or vice versa).
    so besides this distinction, the rest of the proof is valid.
    (correct me if I'm wrong)
    Last edited by Stormey; November 19th 2013 at 01:28 PM.
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    Re: Need help with a proof - subgroups

    [QUOTE=Stormey;804944]
    Let G be a group.
    Let H,K be non-trivial subgroups of G ( \left \{ e \right \} <H,K<G).
    //
    Last edited by Plato; November 19th 2013 at 02:23 PM.
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    Re: Need help with a proof - subgroups

    Quote Originally Posted by Stormey View Post
    Let G be a group.
    Let H,K be non-trivial subgroups of G ( \left \{ e \right \} <H,K<G).
    Prove that H\cup K\neq G.
    I am confused by the notation.
    I would assume that ( \left \{ e \right \} <H,K<G means that each of K~\&~H are proper subgroups.

    If that is the case, then it must be true that neither is a subset of the other.
    So \left( {\exists k \in K} \right)\left[ {k \notin H} \right] \wedge \left( {\exists h \in H} \right)\left[ {h \notin K} \right]

    Now can you argue that hk \in G BUT hk \notin H \wedge hk \notin K.

    Did I miss-understand that notation?
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    Re: Need help with a proof - subgroups

    Quote Originally Posted by Plato View Post
    I am confused by the notation.
    I would assume that ( \left \{ e \right \} <H,K<G means that each of K~\&~H are proper subgroups.

    If that is the case, then it must be true that neither is a subset of the other.
    So \left( {\exists k \in K} \right)\left[ {k \notin H} \right] \wedge \left( {\exists h \in H} \right)\left[ {h \notin K} \right]

    Now can you argue that hk \in G BUT hk \notin H \wedge hk \notin K.

    Did I miss-understand that notation?
    No, you got it. that is indeed the meaning of this notation.

    Why can I argue that \left( {\exists k \in K} \right)\left[ {k \notin H} \right] \wedge \left( {\exists h \in H} \right)\left[ {h \notin K} \right]? how do I justify it?
    Last edited by Stormey; November 19th 2013 at 02:44 PM.
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    Re: Need help with a proof - subgroups

    Quote Originally Posted by Stormey View Post
    No, you got it. that is indeed the meaning of this notation.
    Why can I argue that \left( {\exists k \in K} \right)\left[ {k \notin H} \right] \wedge \left( {\exists h \in H} \right)\left[ {h \notin K} \right]? how do I justify it?
    That is the very meaning of K \not\subset H \wedge H \not\subset K, surely you know that.

    If K \subseteq H \Rightarrow \;K \cup H = H \ne G so neither is the subset of the other.

    If hk \in H \Rightarrow \;hk = h' \Rightarrow \;k = {h^{ - 1}}h' \in H.

    Same for hk \in K \Rightarrow \;h = k'{k^{ - 1}} \in K
    Thanks from Stormey
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    Re: Need help with a proof - subgroups

    ohh...

    So I actually have three cases:

    I: where H=K.
    II: where H\subset K\vee K\subset K.
    III where K \not\subset H \wedge H \not\subset K.

    Thank you, again.
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