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Math Help - A normal sylow 3-subgroup of D6

  1. #1
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    A normal sylow 3-subgroup of D6

    I could not verify the below paragraph in my textbook [Hungerford, p98].

    Let c \in G, G is a group of order 12, and P is the normal (unique) sylow 3-subgroup. Hence G contains only two elements of order 3. If c is one of these, then [G:C_{G}(c)]=1 or 2. Thus C_{G}(c) is a group of order 12 or 6. In either case there is d \in C_{G}(c) of order 2 by Cauchy's theorem. Verify that |cd| = 6.

    In my textbook, D_{6} is such a group that has a normal sylow 3-subgroup.

    D_{6} = \{1, r, r^2, r^3, r^4, r^5, s, sr, sr^2, sr^3, sr^4, sr^5\}.
    What is the normal sylow 3-subgroup of the above D_{6}?
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  2. #2
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    Quote Originally Posted by aliceinwonderland View Post
    I could not verify the below paragraph in my textbook [Hungerford, p98].

    Let c \in G, G is a group of order 12, and P is the normal (unique) sylow 3-subgroup. Hence G contains only two elements of order 3. If c is one of these, then [G:C_{G}(c)]=1 or 2. Thus C_{G}(c) is a group of order 12 or 6. In either case there is d \in C_{G}(c) of order 2 by Cauchy's theorem. Verify that |cd| = 6.
    If a,b\in G are elements that commute (with finite orders) then |ab| = \text{lcm}(|a|,|b|).
    Now |c|=3 and |d|=2. However, cd=dc because by definition d\in C_G(c).
    The rest follows.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    If a,b\in G are elements that commute (with finite orders) then |ab| = \text{lcm}(|a|,|b|).
    unfortunately this is not always true, even if we assume the group is abelian. (for example, think about when a=b) we also need the orders to be coprime, which in this case: |ab|=|a||b|.

    however, it is true that there always exists an element of G of order \text{lcm}(|a|,|b|). this is a good problem which is worth thinking!

    for the second part of aliceinwonderland's question: the Sylow 3-subgroup of D_6 is: <r^2>=\{1,r^2,r^4 \}.
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    Quote Originally Posted by NonCommAlg View Post
    unfortunately this is not always true, even if we assume the group is abelian. (for example, think about when a=b) we also need the orders to be coprime, which in this case: |ab|=|a||b|.
    Yes , what was I thinking! But lucky my argument works because |ab| = |a||b| as you said.

    however, it is true that there always exists an element of G of order \text{lcm}(|a|,|b|). this is a good problem which is worth thinking!
    I posted a solution a year ago, here via Sylow.
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    Quote Originally Posted by ThePerfectHacker View Post

    I posted a solution a year ago, here via Sylow.
    the problem i gave you was slightly different. it was about cyclic subgroups. here is the complete question:

    let G be a group (abelian or non-abelian) and a,b \in G two elements of finite order which commute. show that G has an element of order \text{lcm}(|a|,|b|).
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  6. #6
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    Quote Originally Posted by aliceinwonderland View Post
    Let c \in G, G is a group of order 12, and P is the normal (unique) sylow 3-subgroup. Hence G contains only two elements of order 3. If c is one of these, then [G:C_{G}(c)]=1 or 2. Thus C_{G}(c) is a group of order 12 or 6. In either case there is d \in C_{G}(c) of order 2 by Cauchy's theorem. Verify that |cd| = 6.
    Thanks for all replies.

    I have one last paragraph in my text that I could not understand regarding a nonabelian group of order 12.

    "Let a = cd (in the above quote); then <a> is normal in G and |G/<a>| = 2. Hence there is an element  b \in G such that  b \notin <a>, b \neq e, b^{2} \in <a>, bab^{-1} \in <a>.
    Since G is nonabelian and |a|=6, bab^{-1} = a^{5} = a^{-1} is the only possibility.
    ....
    There are six possibilities for b^{2} \in <a>. b^{2}=a^{2}, b^{2}=a^{4} lead to contradictions.  b^{2}=a or b^{2}=a^{5} imply |b|=12.
    ..........."

    Why the b^{2}=a^{2}, b^{2}=a^{4} lead to contradiction?
    Last edited by aliceinwonderland; December 26th 2008 at 01:15 AM.
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    Quote Originally Posted by aliceinwonderland View Post
    Thanks for all replies.

    I have one last paragraph in my text that I could not understand regarding a nonabelian group of order 12.

    "Let a = cd (in the above quote); then <a> is normal in G and |G/<a>| = 2. Hence there is an element  b \in G such that  b \notin <a>, b \neq e, b^{2} \in <a>, bab^{-1} \in <a>.
    Since G is nonabelian and |a|=6, bab^{-1} = a^{5} = a^{-1} is the only possibility.
    ....
    There are six possibilities for b^{2} \in <a>. b^{2}=a^{2}, b^{2}=a^{4} lead to contradictions.  b^{2}=a or b^{2}=a^{5} imply |b|=12.
    ..........."

    Why the b^{2}=a^{2}, b^{2}=a^{4} lead to contradiction?
    suppose b^2=a^2. then from bab^{-1}=a^{-1}, we get a^3b^{-1}=b^2ab^{-1}=ba^{-1}. thus: a^{-3}=(bab^{-1})^3=ba^3b^{-1}=b^2a^{-1}=a. hence a^4=e, which contradicts |a|=6. do the same for the case b^2=a^4.
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