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Math Help - abstract algebra proof

  1. #1
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    abstract algebra proof

    Can anyone help with the following?
    Prove that G cannot have a subgroup H with the order of H=n-1, where the order of G=n, and n is greater than 2.
    Thanks!
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  2. #2
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    Quote Originally Posted by marches View Post
    Can anyone help with the following?
    Prove that G cannot have a subgroup H with the order of H=n-1, where the order of G=n, and n is greater than 2.
    Thanks!

    Lagrange's Theorem: when n-1\mid n ?

    Tonio
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  3. #3
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    Thanks, but can't use Lagrange's...not to mention, G and H are infinite--I think Lagrange applies to finite groups. Any other suggestions?
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  4. #4
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    Quote Originally Posted by marches View Post
    Thanks, but can't use Lagrange's...not to mention, G and H are infinite--I think Lagrange applies to finite groups. Any other suggestions?

    You gave both G, H FINITE groups! You said the order of H = n-1 and the order of G = n...of course Lagrange applies only in the finite case: the case you gave.

    Tonio
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  5. #5
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    oops--Right,right, I was thinking the n>2 thing was implying infinite order. But still, we were told not to use Lagrange's. Any other ideas?
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  6. #6
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    Quote Originally Posted by marches View Post
    oops--Right,right, I was thinking the n>2 thing was implying infinite order. But still, we were told not to use Lagrange's. Any other ideas?
    We have that G-H=\{m\}. Now, clearly m\ne e. Now, since n>2 we see that |H|>1 and so there exists some h\in H-\{e\}. Clearly mh\in G but mh\notin H since if it were we'd have that mh=h',\text{ }h'\in H and so m=h'h^{-1}\implies m\in H. Thus, mh\notin H\implies mh=m\implies h=e. This contradicts our choice of h. The conclusion follows.
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