
group
Hello
1. G is a group of order $\displaystyle p^n$, with p prime number and n natural number, which acts on a finite set of cardinal X not divisible by p. Show that there is some element x in X such that gx = x for all g in G.
2. If p and q are primes with p <q. Prove that every group G of order pq has only one subgroup of order q normal in G. If q is not congruent to 1 modulo p, show that G is abelian and cyclic.
Thanks, I hope some help

Re: group
Hi, resident911.
Some tips:
Problem #1:
Define
$\displaystyle X^{G}=\{x\in X : gx=x \forall g\in G\}.$
Since G is a pgroup, a consequence of the orbitstabilizer theorem is that
$\displaystyle X\equiv X^{G}$ mod p.
Hopefully this looks familiar from class or a book. If not I can give more details later.
Now use our assumption on the order of X and see what you can do.
Problem #2:
I would suggest looking at the Sylow theorems and seeing where they take you.
Give it some thought and I'm sure you'll get it!
Let me know if there's still confusion. Good luck!

Re: group
one doesn't need the sylow theorems for problem #2:
by cauchy's theorem, G has a subgroup H of order p, and a subgroup K of order q. since H∩K = {e}, G = HK.
now [G:K] = p, which means we can let G act on the coset space G/K to get a homomorphism φ:G→S_{p}.
consider φ(K). for any element x in φ(K), we have x divides q. thus either K is in the kernel of φ, or φ(K) ≅ K.
but S_{p} contains no elements of order q, since q does not divide p! (because p < q). so K lies in the kernel.
so for any k in K, and g in G, k(gK) = gK, so g^{1}kgK = K, that is: g^{1}kg is in K.
thus K is normal in G.
this means that for any h in H (which is certainly an element of G) hkh^{1} is in K.
in other words, every h in H induces an action on K by conjugation h.k = hkh^{1}.
this gives a homomorphism from H to Aut(K), and since K is cyclic of order q, Aut(K) = q1.
suppose h in H is a generator. then the automorphism k → hkh^{1} has order dividing p, and q1.
if p does not divide q1, then the automorphism k → hkh^{1} must be the identity automorphism:
i.e.; hkh^{1} = k, for all k in K. in particular, if k is a generator for K, then h and k commute.
but G = HK, so we have for any g in G, g = h^{m}k^{n}, for some 0 ≤ m < p, 0 ≤ n < q.
thus gg' = (h^{m}k^{n})(h^{m'}k^{n'}) = (h^{m+m'})(k^{n+n'})
= (h^{m'+m})(k^{n'+n}) = (h^{m'}k^{n'})(h^{m}k^{n}) = g'g, G is abelian.
but if G is abelian, then H is also normal in G, whence G = H x K ≅ Z_{p} x Z_{q} ≅ Z_{pq}, since gcd(p,q) = 1
(one can also show directly that for a generator h of H, and generator k of K, the element hk has order pq = lcm(p,q)).
*****************
to show that a group of order pq where p divides q1 need not be abelian (let alone cyclic), let p = 2, and q be any odd prime, and consider the dihedral group of order 2q (for q = 3, this is isomorphic to S_{3}).
*****************
in all fairness, a proof using the sylow theorems *would* be shorter in showing K is normal in G. but that would not tell one how to show that HK is abelian.

Re: group
Exercise 2 I stay clear, thanks Deveno.
In Exercise 1 as I can use the hypothesis on the order of X? To complete the exercise? I have no very clear that

Re: group
you want to prove that X^{G} ≥ 1, right? this is the same as proving X^{G} ≠ 0.
if a = b (mod p) this means:
a = b + kp.
suppose p divides b. then p divides b + kp, and thus p divides a.
now if we already know p does NOT divide a, then p cannot possibly divide b, either. in particular, b cannot be 0.

Re: group
hi
understood the exercise, thank you very much.