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Math Help - N(H)/C(H) question

  1. #1
    nhk
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    N(H)/C(H) question

    Assume G is a group of order 48. If there is only one 3-Sylow subgroup, K, of G, what are the possible orders for C(K)? Does this guarantee G has an element of order 6? Why or why not.
    Here is what I have for my answer:
    Since |G|=48=2^4*3. 1 Sylow 3 subgroup means that this subgroup is normal in G, this group order is 3. Since K is a subgroup of C(K) and C(K) is a subgroup of G, by lagrange theroem the possible orders for C(K) are 3,6,12,24,48. Since K is normal in G, then N(K)=G, and |N(K)|=48. So |N(K)/C(K)| and the possible orders of C(K) are 3,6,12,24, and 48.
    G is not guaranteed an element of order 6 because 6 is not a prime number that divides the order of G.
    Is this right? Any comments would be great!
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    Quote Originally Posted by nhk View Post
    Assume G is a group of order 48. If there is only one 3-Sylow subgroup, K, of G, what are the possible orders for C(K)? Does this guarantee G has an element of order 6? Why or why not.
    Here is what I have for my answer:
    Since |G|=48=2^4*3. 1 Sylow 3 subgroup means that this subgroup is normal in G, this group order is 3. Since K is a subgroup of C(K) and C(K) is a subgroup of G, by lagrange theroem the possible orders for C(K) are 3,6,12,24,48. Since K is normal in G, then N(K)=G, and |N(K)|=48. So |N(K)/C(K)| and the possible orders of C(K) are 3,6,12,24, and 48.
    G is not guaranteed an element of order 6 because 6 is not a prime number that divides the order of G.
    Is this right? Any comments would be great!

    What is C(K) for you? The center of the subgroup K or the centralizer in G of K?

    Tonio
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  3. #3
    nhk
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    centralizer in G of K.
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    Quote Originally Posted by nhk View Post
    centralizer in G of K.

    Ok then: we know that N(K)/C(K)\cong S\leq Aut(K) , and since |Aut(K)|=2 that doesn't leave many options, does it?
    So it must be \frac{48}{|C(K)|}=|N(K)/C(K)|\in\{1,2\}\Longrightarrow |C(K)|=24,\,48.

    In both cases this means that if K=<k> then there is an element x\in G of order 2 s.t. xk=kx , and then ord(xk)=ord(x)ord(k)=6 ...

    Tonio
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