# Sylow's theorem

• Feb 18th 2007, 05:36 AM
jedoob
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?
• Feb 18th 2007, 05:47 AM
ThePerfectHacker
Quote:

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.
• Feb 18th 2007, 06:02 AM
jedoob
Quote:

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)
• Feb 18th 2007, 06:13 AM
ThePerfectHacker
Quote:

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.
• Feb 18th 2007, 06:31 AM
jedoob
Thanks.

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