# Sylow's Theorem and Recognition Criterion for Groups of Order pq

• mathusers
In summary, using Sylow's theorem, we can show that a finite group G with |G| = pq has a normal subgroup K \cong G. The Recognition Criterion can then be used to show that G \cong C_p \rtimes_h C_q for some homomorphism h:C_q \rightarrow Aut(C_p). To describe all homomorphisms h:C_5 \rightarrow Aut(C_7), we can use the explicit description of Aut(C_{13}) as the set of all powers of x, where x is a generator of C_{13}. Using this, we can show that there is only one group of order 39, up to isomorphism.
mathusers
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 :)

(i) is trivial as stated. Did you mean to say K=~C_p?

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no the first part was meant as it was shown. but do you have an idea for the other parts please? i don't have a strong enough idea right now to solve them but if you provide a clue then i'll attempt it myself. thnx a lot :)

ok i managed to get part (i)

G a Sylow q-subgroup, we let x be the number of these subgroups. As a result, x|p so x=1 or x=p and x = 1 (mod q) so x=1 or p.
x can't be p because p = 1(mod q) which contradicts the fact that q>p. Therefore, x=1 and so its Sylow q-subgroup has to be a normal subgroup.

any help on the other parts would be greatly appreciated :)

mathusers said:
ok i managed to get part (i)

G a Sylow q-subgroup, we let x be the number of these subgroups. As a result, x|p so x=1 or x=p and x = 1 (mod q) so x=1 or p.
x can't be p because p = 1(mod q) which contradicts the fact that q>p. Therefore, x=1 and so its Sylow q-subgroup has to be a normal subgroup.

You might as well write $n_q$ rather than $x$, and the argument could be clearer:
Sylow gives:
$$n_q | p \rightarrow n_q \in \{1,p\} \rightarrow n_q \leq p$$
then
$$p < q \rightarrow n_q \leq q[/itex] thus [tex]n_q \equiv 1 (\rm {mod} q) \rightarrow n_q = 1$$

I'm unfamiliar with the expression 'recognition criterion' but the second part is similar to this problem -- I'm not sure if you got that one.

for part (ii) i can't find any decent material to learn off and wikipedia doesn't seem to have much on the recognition theorem. any help on this is therefore appreciated aswell.

for part (iv)
mathusers said:
(iv) Describe explicitly all homomorphisms $h:C_3 \rightarrow Aut(C_{13})$.

could you help me on this part please? how can you show the homomorphisms explicitly?

mathusers said:
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. So there is only 1 subgroup of order 3.
Now since 13 and 3 are co prime, the intersection of these 2 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?

"Recognition Criterion" not "Recognition Theorem".

Regarding:
$$C_3 \rightarrow Aut(C_{13})$$
Can you describe $Aut(C_{13})$?
-

$n_3=13$ also is 1 mod 3, and divides 13...

NateTG said:
"Recognition Criterion" not "Recognition Theorem".
still can't manage to find much about that I am afraid? any more ideas on that?

NateTG said:
Regarding:
$$C_3 \rightarrow Aut(C_{13})$$
Can you describe $Aut(C_{13})$?

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?

NateTG said:
$n_3=13$ also is 1 mod 3, and divides 13...

ok so if n_3 = 13 is also a possibility then there are only 2 subgroups of of order 3.. how does this change the final answer though?

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## 1. What is Sylow's Theorem?

Sylow's Theorem is a fundamental result in group theory that gives a powerful tool for understanding the structure of finite groups. It states that if a prime number p divides the order of a finite group G, then G contains a subgroup of order p^k for some positive integer k. This subgroup is called a Sylow p-subgroup.

## 2. What is the Recognition Criterion for Groups of Order pq?

The Recognition Criterion for Groups of Order pq is a corollary of Sylow's Theorem that provides a necessary and sufficient condition for a group of order pq to be isomorphic to a specific group. It states that if a group G has order pq, where p and q are distinct primes and p > q, and if G contains a subgroup of order q, then G is isomorphic to the group of order pq.

## 3. How can Sylow's Theorem be used to classify groups of order pq?

Sylow's Theorem can be used to classify groups of order pq by identifying all possible subgroups of order p and q in a given group G. By the Recognition Criterion, if G contains a subgroup of order q, then it must be isomorphic to the group of order pq. Similarly, if G contains a subgroup of order p, then it must be isomorphic to the group of order pq. By identifying all possible subgroups of order p and q, we can determine all possible structures of groups of order pq.

## 4. Are there any limitations to Sylow's Theorem and the Recognition Criterion?

Yes, there are limitations to Sylow's Theorem and the Recognition Criterion. They only apply to finite groups, and even then, they may not always be sufficient to determine the entire structure of a group. Additionally, the criteria for a group of order pq to be isomorphic to the group of order pq can be quite restrictive and may not apply to all groups of that order.

## 5. Can Sylow's Theorem and the Recognition Criterion be applied to groups of order other than pq?

Yes, Sylow's Theorem and the Recognition Criterion can be applied to groups of order other than pq. In fact, Sylow's Theorem can be generalized to apply to groups of any order, not just those of the form pq. However, the Recognition Criterion is specific to groups of order pq and does not have a generalization for other orders.

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