# Thread: Sylow's theorem

1. ## Sylow's theorem

Hi,

Please can someone help me with the following problems:

a and b are distinct primes such that a<b, and H is a finite group such that
l H l = ab

1) How can I use Sylows theorem to show that H has a normal subgroup K with K being isomorphic to Cp (C is the cyclic group).

2) Describe explicitly all homomorphisms h: C5 --> Aut(C7), and from this, describe all groups of order 35. How many such subgroups are there?

3) Same as 2) but this time h: C3 --> Aut(C13), and from this, describe all groups of order 39. Again, how many such subgroups are there?

2. Originally Posted by jedoob

1) How can I use Sylows theorem to show that H has a normal subgroup K with K being isomorphic to Cp (C is the cyclic group).
This statement is not true!

The theorem says if |G|=pq with q>p, and q is not congruent to 1 modulo p then |G| is a cyclic group.
However, when it is congruent to 1 modulo p we cannot conlude that.

However we can show that |G| is not a simple group.
-Proof

Let |G|=pq and p>q.
Now by Sylow's First Theorem there exists a Sylow p-subgroup. Let S(p) denote the number of Sylow p-subgroups. Then we know that by Sylow's Third Theorem:
S(p) = 1 (mod p) and S(p)|(pq).
Thus, the statement S(p)|(pq) states that,
S(p)=1,p,q,pq
By checking each one we see that,
p,q,pq do not work in the congruence (because p > q).
Thus, S(p)=1.
Now, any inner automorphism under any element maps a Sylow p-subgroup into a Sylow p-subgroup. Since the number of Sylow p-subgroups is fixed at one, it means that the Sylow p-subgroup remains invariant under conjugation, i.e. it is a normal subgroup of G, and hence G is not simple.

3. Originally Posted by jedoob
1) How can I use Sylows theorem to show that H has a normal subgroup K with K being isomorphic to Cp (C is the cyclic group).
sorry i meant K being isomorphic to Cb (not Cp as quoted)

4. Originally Posted by jedoob
sorry i meant K being isomorphic to Cb (not Cp as quoted)
Simple (get the pun) we already shown that there exists a Sylow p-subgroup (p is prime), and there is only one group up to isomorphism of order p, that is, the cyclic group. Proof complete.

5. Thanks.

Could you possibly help me with parts 2 and 3 as well?