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

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

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

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

    Let $\displaystyle k\in K$.

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

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

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

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

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

    But then:

    $\displaystyle h\cdot k=m$

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

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

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

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


    any help would be greatly appreciated!
    Last edited by Stormey; Nov 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 $\displaystyle G$ be a group.
    Let $\displaystyle H,K$ be non-trivial subgroups of $\displaystyle G$ ($\displaystyle \left \{ e \right \} <H,K<G$).
    Prove that $\displaystyle H\cup K\neq G$.

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

    If $\displaystyle H\neq K$, without loss of generality, suppose that $\displaystyle K<H$.
    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; Nov 19th 2013 at 01:28 PM.
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    Re: Need help with a proof - subgroups

    [QUOTE=Stormey;804944]
    Let $\displaystyle G$ be a group.
    Let $\displaystyle H,K$ be non-trivial subgroups of $\displaystyle G$ ($\displaystyle \left \{ e \right \} <H,K<G$).
    //
    Last edited by Plato; Nov 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 $\displaystyle G$ be a group.
    Let $\displaystyle H,K$ be non-trivial subgroups of $\displaystyle G$ ($\displaystyle \left \{ e \right \} <H,K<G$).
    Prove that $\displaystyle H\cup K\neq G$.
    I am confused by the notation.
    I would assume that ($\displaystyle \left \{ e \right \} <H,K<G$ means that each of $\displaystyle K~\&~H$ are proper subgroups.

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

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

    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 ($\displaystyle \left \{ e \right \} <H,K<G$ means that each of $\displaystyle K~\&~H$ are proper subgroups.

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

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

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

    Why can I argue that $\displaystyle \left( {\exists k \in K} \right)\left[ {k \notin H} \right] \wedge \left( {\exists h \in H} \right)\left[ {h \notin K} \right]$? how do I justify it?
    Last edited by Stormey; Nov 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 $\displaystyle \left( {\exists k \in K} \right)\left[ {k \notin H} \right] \wedge \left( {\exists h \in H} \right)\left[ {h \notin K} \right]$? how do I justify it?
    That is the very meaning of $\displaystyle K \not\subset H \wedge H \not\subset K$, surely you know that.

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

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

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

    ohh...

    So I actually have three cases:

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

    Thank you, again.
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