Results 1 to 4 of 4

- Jun 1st 2010, 12:15 PM #1

- Joined
- Apr 2010
- Posts
- 43

## 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!

- Jun 1st 2010, 07:34 PM #2

- Joined
- Oct 2009
- Posts
- 4,261
- Thanks
- 2

- Jun 1st 2010, 08:20 PM #3

- Joined
- Apr 2010
- Posts
- 43

- Jun 1st 2010, 08:44 PM #4

- Joined
- Oct 2009
- Posts
- 4,261
- Thanks
- 2