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Thread: abstract algebra

  1. #1
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    abstract algebra

    Prove that if $\displaystyle H$ is a normal subgroup of $\displaystyle G$ of prime index $\displaystyle p$ then for all subgroups $\displaystyle K$ of $\displaystyle G$ either
    (i) $\displaystyle K$ is a subgroup of $\displaystyle H$, or
    (ii) $\displaystyle G = HK$ and $\displaystyle |K : K$ intersect $\displaystyle H| = p$

    I tried to do this in 2 cases. The first case is to let $\displaystyle K$ be a subest of $\displaystyle H$, then since $\displaystyle K$ is a subgroup of $\displaystyle G$, it is a subgroup of $\displaystyle H$. For case 2, I let $\displaystyle K$ not contained in $\displaystyle H$, then try to show $\displaystyle G = HK$ and $\displaystyle |K : K$ intersect $\displaystyle H| = p$, but don't know how... please help, thank you.
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  2. #2
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    Let $\displaystyle \pi:G\to G/H$ be the quotient map and consider $\displaystyle \pi(K)$. (It is a subgroup of the simple group G/H.)
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  3. #3
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    Quote Originally Posted by Opalg View Post
    Let $\displaystyle \pi:G\to G/H$ be the quotient map and consider $\displaystyle \pi(K)$. (It is a subgroup of the simple group G/H.)
    I don't get it...
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    Quote Originally Posted by dori1123 View Post
    I don't get it...
    What Opalg means is this:

    The group $\displaystyle G/H$ has prime order, therefore, it only has no proper non-trivial subgroup. The natural projection $\displaystyle \pi : G\to G/H$ defined by $\displaystyle \pi (a) = aH$ is a homomorphism. Thus, by the property of homomorphism, if $\displaystyle K$ is a subgroup of $\displaystyle G$ then $\displaystyle \pi (K)$ is a subgroup of $\displaystyle G/H$. This means that $\displaystyle \pi (K) = G/H$ or $\displaystyle \pi (K) = \{ H \}$. We have two cases, (i) $\displaystyle \pi (K) = G/H$ (ii) $\displaystyle \pi (K) = \{ H\}$. Now consider each case.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    if $\displaystyle K$ is a subgroup of $\displaystyle G$ then $\displaystyle \pi (K)$ is a subgroup of $\displaystyle G/H$. This means that $\displaystyle \pi (K) = G/H$ or $\displaystyle \pi (K) = \{ H \}$. We have two cases, (i) $\displaystyle \pi (K) = G/H$ (ii) $\displaystyle \pi (K) = \{ H\}$. Now consider each case.
    How do you know $\displaystyle \pi (K) = G/H$ or $\displaystyle \pi (K) = \{ H \}$?
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  6. #6
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    Quote Originally Posted by dori1123 View Post
    How do you know $\displaystyle \pi (K) = G/H$ or $\displaystyle \pi (K) = \{ H \}$?
    Because $\displaystyle G/H$ has only non-proper and trivial subgroups. That is $\displaystyle G/H$ - the full group itself. And $\displaystyle \{ H \}$ - the set containing the identity element.
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