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Math Help - Plz help me on this.....

  1. #1
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    Lightbulb Plz help me on this.....

    Q1:- Prove that a subgroup H of G is normal subgroup of G,

    if and only if


    product of any two right cosets of H in G is again a right coset of H in G.


    Plz help me on this.....-img_20120101_062550.jpg



    dnt know wether i can write this or not....???



    Q2:- If H is a non-empty subset of G such that Hx = HxH for every xЄG,

    Then Hx = xH.



    Q3:- If H and K are two normal subgroups of a finite group G such that O(H) and O(K) are
    co-prime.

    Then show that hk = kh for every hЄH, kЄK .
    Last edited by Aman3230; March 29th 2014 at 05:44 AM.
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  2. #2
    Senior Member vincisonfire's Avatar
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    Re: Plz help me on this.....

    In general, (xh)^2 = xhxh \neq x^2h^2 and hence you first to last step doesn't work.
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  3. #3
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    Re: Plz help me on this.....

    Hi,
    Some preliminary observations:
    Let $G$ be any group and $A, B$ subsets of $G$. The product set is defined by $AB=\{ab:a\in A \text{ and }b\in B\}$. If A or B is a singleton, write $aB$ or $Ab$ for $AB$. Let $C$ be another subset, easily: if $A\subset B$, then $AC\subset BC$ (also you can multiply by C on the left). So if $A=B$, $AC=BC$.

    1. Let $H$ be a subgroup of $G$. Then $H$ is normal in $G$ if and only if $xH=Hx$ for all $x\in G$.
    a) Assume $H$ is normal in $G$ and $x\in G$. Then $x^{-1}Hx=H$. Thus $xx^{-1}Hx=xH$ and so $Hx=xH$.
    b) Assume $Hx=xH$ for all $x\in G$. Let $x\in G$. Then $x^{-1}Hx=x^{-1}xH=H$. So $H$ is normal in $G$.

    2. Assume $H$ is a non-empty subset of G with $Hx=HxH$ for all $x\in G$. Then $xH=Hx$ for all $x\in G$.

    I can't work this problem (I may be overlooking something that is obvious). Some comments, though ( 1 is the identity of $G$):

    Suppose $1\in H$. Let $x\in G$. Then $1xH\subset HxH=Hx$ and so $xH\subset Hx$. So $H=x^{-1}xH\subset x^{-1}Hx$ for any x. Choosing $x^{-1}$, I get $H\subset (x^{-1})^{-1}Hx^{-1}$ and so $H\subset xHx^{-1}$ and then $x^{-1}Hx\subset H$. Thus $H=x^{-1}Hx$. By the same argument as in 1a), I get $Hx=xH$.

    Next $H1=H1H$ and so $H=HH$; i.e $H$ is closed under multiplication. So if $G$ is finite or even if $H$ contains an element of finite order, then $1\in H$ and I'm done. So a counter example must be an infinite non-abelian group with no elements of finite order in $H$. I haven't been able to come up with such an example.

    3. Hints: Determine the subgroup $H\cap K$. Next for $h\in H, k\in K$, examine the element $h^{-1}k^{-1}hk$.
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  4. #4
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    Re: Plz help me on this.....

    Quote Originally Posted by johng View Post
    Hi,
    Some preliminary observations:
    Let $G$ be any group and $A, B$ subsets of $G$. The product set is defined by $AB=\{ab:a\in A \text{ and }b\in B\}$. If A or B is a singleton, write $aB$ or $Ab$ for $AB$. Let $C$ be another subset, easily: if $A\subset B$, then $AC\subset BC$ (also you can multiply by C on the left). So if $A=B$, $AC=BC$.

    1. Let $H$ be a subgroup of $G$. Then $H$ is normal in $G$ if and only if $xH=Hx$ for all $x\in G$.
    a) Assume $H$ is normal in $G$ and $x\in G$. Then $x^{-1}Hx=H$. Thus $xx^{-1}Hx=xH$ and so $Hx=xH$.
    b) Assume $Hx=xH$ for all $x\in G$. Let $x\in G$. Then $x^{-1}Hx=x^{-1}xH=H$. So $H$ is normal in $G$.

    2. Assume $H$ is a non-empty subset of G with $Hx=HxH$ for all $x\in G$. Then $xH=Hx$ for all $x\in G$.

    I can't work this problem (I may be overlooking something that is obvious). Some comments, though ( 1 is the identity of $G$):

    Suppose $1\in H$. Let $x\in G$. Then $1xH\subset HxH=Hx$ and so $xH\subset Hx$. So $H=x^{-1}xH\subset x^{-1}Hx$ for any x. Choosing $x^{-1}$, I get $H\subset (x^{-1})^{-1}Hx^{-1}$ and so $H\subset xHx^{-1}$ and then $x^{-1}Hx\subset H$. Thus $H=x^{-1}Hx$. By the same argument as in 1a), I get $Hx=xH$.

    Next $H1=H1H$ and so $H=HH$; i.e $H$ is closed under multiplication. So if $G$ is finite or even if $H$ contains an element of finite order, then $1\in H$ and I'm done. So a counter example must be an infinite non-abelian group with no elements of finite order in $H$. I haven't been able to come up with such an example.

    3. Hints: Determine the subgroup $H\cap K$. Next for $h\in H, k\in K$, examine the element $h^{-1}k^{-1}hk$.


    Plz help me on this.....-11.jpg
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  5. #5
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    Re: Plz help me on this.....

    Hi again,

    Your problem stated that H is a non-empty subset of G, not necessarily a subgroup.
    From H=HxHx-1, I don't see how you conclude that xhx-1 is in H without assuming that 1 is an element of H.
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  6. #6
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    Re: Plz help me on this.....

    No, that's not right. Suppose Ha = Hb. This means that:

    ha = h'b, for some h,h' in H. Yes, we CAN multiply both sides by b-1, to get:

    h(ab-1) = h', but....

    from this, all we can conclude is:

    ab-1 in h-1H, and there is NO guarantee that this equals H (if H is NOT a subgroup, then it probably ISN'T).

    I think there must be a typo in the problem, because subsets H have different properties than subgroups.
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