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Math Help - two tricky group problems

  1. #1
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    two tricky group problems

    (1)

    Prove that if N is a normal subgroup of the finite group G and (|N|,|G:N|)=1 then N is the unique subgroup of G of order |N|.
    (2)

    Let K be a subgroup of some group (possibly infinite) G, and let H be a subgroup of K. Prove that |G:H|=|G:K|\cdot|K:H|.
    Regarding notation, |G:H| means the number of left cosets of H in G, that is, the number of elements in the set \{gH:g\in G\}. Also, |N| is just the order of N, that is, the number of elements in N. Finally, (a,b) denotes the greatest common divisor of the integers a,b.

    I could probably get these with enough time, but I'm fresh out of it. Any hints (or heck, even just complete solutions) would be much appreciated!
    Last edited by hatsoff; September 21st 2010 at 09:11 PM.
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  2. #2
    Senior Member roninpro's Avatar
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    For problem 1, since G is finite, we have |G:N|=|G|/|N|, so the condition can be rephrased as \gcd(|N|,|G|/|N|)=1. I would try to prove by contradiction: suppose there is a another subgroup N' such that |N'|=|N|. Then I would consider the quotient group G/N and try mapping N' into it, using the canonical homomorphism.

    For problem 2, we can assume that |G:K| and |K:H| are finite. Let the collection of left cosets of K in G be \{a_iK\ |\ i=1,2,\ldots r\} and the left cosets of H in K be \{b_jH\ |\ j=1,2,\ldots s\}. It would suffice to show that the set \{a_ib_jH\ |\ i=1,2,\ldots r; j=1,2,\ldots s\} is the collection of left cosets of H in G.

    Can you give that a shot?
    Last edited by roninpro; September 21st 2010 at 11:13 PM.
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  3. #3
    MHF Contributor Swlabr's Avatar
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    You may not have it in your group theory arsenal yet, but if you do,

    Question 1. follows easily from Sylow's Theorem.
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  4. #4
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    Thanks for the hints guys! I was able to get (1). However, problem (2) is still giving me trouble...

    Quote Originally Posted by roninpro View Post
    For problem 2, we can assume that |G:K| and |K:H| are finite. Let the collection of left cosets of K in G be \{a_iK\ |\ i=1,2,\ldots r\} and the left cosets of H in K be \{b_jH\ |\ j=1,2,\ldots s\}. It would suffice to show that the set \{a_ib_jH\ |\ i=1,2,\ldots r; j=1,2,\ldots s\} is the collection of left cosets of H in G.
    That's the approach I initially planned, but I just haven't had any luck getting the proof to fall together. Ordinarily I'd take a day or two away from it, and come back to it later with fresh eyes. Unfortunately, I'm nearly out of time, with only a couple hours to go.
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  5. #5
    Senior Member roninpro's Avatar
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    The proof has two parts: you must show that (1) all of the cosets are distinct and (2) every left coset of H in G can be written in the form a_ib_jH.

    For (1), what happens if a_ib_jH=a_\alpha b_\beta H?

    For (2), what would it take for gH=a_ib_iH (where g\in G)?
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