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Thread: subgroup

  1. #1
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    subgroup

    Let H be a group. Let M be a normal subgroup of H. Let K be any subgroup of H. Let MK={mk: m is an element of M and k is an element if K}
    a)Prove: MK is a subgroup of H.
    b)Suppose that M intersects K={I}. Let k, k' are elements of K. Prove Mk=M k' if and only if k= k'. conclude that |MK|=|M||K|
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  2. #2
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    Quote Originally Posted by apple2009 View Post
    Let H be a group. Let M be a normal subgroup of H. Let K be any subgroup of H. Let MK={mk: m is an element of M and k is an element if K}
    a)Prove: MK is a subgroup of H.
    b)Suppose that M intersects K={I}. Let k, k' are elements of K. Prove Mk=M k' if and only if k= k'. conclude that |MK|=|M||K|
    For (a), it would be helpful to use the following lemma.

    Lemma. If M and K are subgroups of a group H, MK is a subgroup of H if and only if MK = KM.

    Now you need to show that $\displaystyle MK \subseteq KM$ and $\displaystyle KM \subseteq MK$ in order to apply the above lemma. I'll show $\displaystyle MK \subseteq KM$ and I'll leave it to you to show that the reverse inclusion.

    Let $\displaystyle m \in M, k \in K$ such that $\displaystyle mk \in MK$. Since M is a normal subgroup of H, $\displaystyle mk = k(k^{-1}mk) \in KM$. Thus $\displaystyle MK \subseteq KM$.

    (b) You might need to use this.
    $\displaystyle |MK| = \frac{(|M|)(|K|)}{|M \cap K|}$
    Last edited by aliceinwonderland; Nov 8th 2009 at 10:13 AM.
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