# Math Help - Sylow's Theorem

1. ## Sylow's Theorem

Let p,q be distinct primes with q < p and let G be a finite group with |G| = pq.

(i) Use sylow's theorem to show that G has a normal subgroup K with $K \cong G$

(ii) Use the Recogition Criterion to show $G \cong C_p \rtimes_h C_q$ for some homomorphism $h:C_q \rightarrow Aut(C_p)$

(iii) Describle explicitly all homomorphisms $h:C_5 \rightarrow Aut(C_7)$. Hence describe all groups of order 35. How many such subgroups are there?

(iv) Describe explicitly all homomorphisms $h:C_3 \rightarrow Aut(C_{13})$. Hence describe all groups of order 39. How many such groups are there, up to isomorphism?

any help is highly appreciated as usual. i will attempt the rest myself once i have good idea. thnx a lot.

2. Originally Posted by joanne_q
Let p,q be distinct primes with q < p and let G be a finite group with |G| = pq.

(i) Use sylow's theorem to show that G has a normal subgroup K
It has a Sylow q-subgroup, let x be the number of these subgroups. Then x|p so x=1 or x=p and x = 1 (mod q) so x=1 or p. It cannot be x=p because then p = 1(mod q) which is impossible since q>p. Thus, x=1 and that means its Sylow q-subgroup must be a normal subgroup.

3. for part (ii) i cant find any decent material to learn off and wikipedia doesnt seem to have much on the recognition theorem. any help on this is therefore appreciated aswell.

for part (iv)
Originally Posted by joanne_q
(iv) Describe explicitly all homomorphisms $h:C_3 \rightarrow Aut(C_{13})$.
ok so $Aut(C_{13})$ is:
$C_{13} = {1,x,x^2,x^3,x^4,..........,x^{12}}$

so $Aut(C_{13}) = {\phi_1,\phi_2,\phi_3,\phi_4,\phi_5,\phi_6,\phi_7, \phi_8,\phi_9,\phi_{10},\phi_{11},\phi_{12}}$ right? so now what would be the next step?

Originally Posted by joanne_q
Hence describe all groups of order 39. How many such groups are there, up to isomorphism?
ok for this part this is what i got:
h is a group of order 39 = 13 x 3.
$n_{13}$ must divide 3. and $n_{13} = 1(mod 13)$. The only value satisfying these constraints is 1. So there is only 1 subgroup of order 13.
Similarly, $n_3$ must divide 13. and $n_3 = 1(mod 3)$. The only value satisfying these constraints is 1 & 3. So there is only 2 subgroups of order 3.
Now since 13 and 3 are co prime, the intersection of these 3 subgroups is trivial and thus there is only 1 group of order 39 upto isomorphism...

is this done correct? please verify and correct if necessary?