1. ## 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!

2. Originally Posted by nhk
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

3. centralizer in G of K.

4. Originally Posted by nhk
centralizer in G of K.

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

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

Tonio