1. Hi,

Can the same semidirect product represent different non-isomorphic groups?

For this seems to be the case in a group of order 130 for instance. This groups has 4 non-isomorphic cases: one cyclic and the others non abelian.

In these 3 others, the subgroup of order 65 (say H65) is cyclic and normal while the subgroup of order 2 (say H2) is cyclic and not normal.

Furthermore, their intersection is trivial.

And noting that H65H2 is a subgroup of G and that |H65H2|=|H65||H2|/1=|G|, then G = H65H2, hence G is the semidirect product of H65 and H2.

Hence, the semidirect product of Z65xZ2 seems to represent 3 different non-isomrphic cases. Is that correct?

Another thing: the dihedral group D65 can be viewed as a semidirect product of Z65xZ2. But is D65 isomorphic to Z65xZ2? Because it is definitely wrong that D65 represents 3 non-somorphic cases.

So, what I beleive is that the semidirect product Z65xZ2 indeed represents 3 different non-isomorphic cases, one of which is D65.

Am I correct?

2.

3. First, sorry for not getting back to you regarding your paper. I've been pretty busy.

Second, yeah, you can come up with different semidirect products using the same groups. The question is basically how one of the groups "acts" on the other. So, for example, all four groups of order 130 appear as semidirect products. Let f: C<sub>2</sub> -> Aut(C<sub>65</sub>) be any homomorphism. Then we can a semidirect product structure on C<sub>2</sub> x C<sub>65</sub> by:

(a,b)+(c,d) = (a+c,b+[f(a)](d))

What are the possible values for f? Well, note that an automorphism of C<sub>65</sub> is determined by the image of a generator, say 1 (viewing C<sub>65</sub> as Z/65Z). 1 must be sent to another generator in order for this to be an automorphism, so 1 can be sent to any integer relatively prime to 65. (Some basic number theory shows that any integer n relatively prime to 65 gives rise to a (nonunique) representation sn + t65 = 1, and so 1 is in the group generated by n modulo 65, and hence it's the whole group.) So let's say f(1) = n, (n,65)=1. Then for any other integer m, f(m) = mf(1) = mn, and so the automorphism is just multiplication by n. Thus, if we apply the automorphism twice, we're multiplying by n<sup>2</sup>. We're looking for automorphisms of order 2, so we're looking for integers n such that n<sup>2</sup> = 1 modulo 65. You can check that there are four possibilities: 1, 14, 51, and 63. So the automorphisms:

f(n) = n
f(n) = 14n
f(n) = 51n
f(n) = 63n

are all the possible ways of generating a semidirect product of C<sub>2</sub> and C<sub>65</sub> with C<sub>65</sub> being the normal subgroup.

4. Could you tell me what field of math the above stuff is dealing with?

5. Thanks very much. Don't worry about the paper, I already gave it to my professor as I told you, who confirmed everything he read till now. Actually, as I was reviewing it, I think I complicated many things for nothing. But whatever.

I beleive I understand everything you said except this

We're looking for automorphisms of order 2
Why?

Also, I beleive the last number is 64, not 63.

Thanks a lot!

6. Could you tell me what field of math the above stuff is dealing with?
Group theory. A very charming branch of Abstract Algebra and my favorite course.

The idea is that a field like Analysis for example always deals with distances, which makes it a bit "static" branch of pure mathematics. But Algebra always deals with operations, and that's why it's very dynamic. Your groups are alive!

But I love Analysis too. What I don't really like is things like "series solutions for differential equations", "numerical analysis"...etc all of that is quite boring for me at least

7. Also,

Let f: C2 -> Aut(C65) be any homomorphism
Why not f: C65-> Aut(C2)? Is that because C65 is the normal subgroup? Or is it arbitrary?

8. You're right, I meant 64. We're looking for automorphisms of order 2 (or order dividing 2) because those are the only possible images of 1 (as an element of Z/2Z) in any group, and any homomorphism of Z/2Z to some group is determined by the image of 1. We could look at f: Z/65Z -> Aut(Z/2Z), but note... Aut(Z/2Z) only has one element! So this will just give us the usual direct product.

In general, if you have two groups H, N and a homomorphism f: H -> Aut(N), you can construct a semidirect product structure on H x N via:

(h,n)(g,m) = (hg,n f(m))

And you can also see that any semidirect product comes from this--it turns out that f is basically just conjugation by H, in the sense that:

(h,e)(e,n)(h<sup>-1</sup>,e) = (e,f(n))

--------------------------

Chemboy: it's my understanding group theory is very useful in chemistry. One of the original applications of group theory is to symmetries of geometric objects, which is important if you're looking at, say, the structure of crystals.

9. Ah, I understand. So the 4 non isomorphic groups have the following binary operations:

(a,b)+(c,d) = (a+c,b+d)
(a,b)+(c,d) = (a+c,b+14d)
(a,b)+(c,d) = (a+c,b+51d)
(a,b)+(c,d) = (a+c,b+64d)

The first case is obviously C130... mmm what about the others?
The problem is that b can be anything in C65, so I can't see the restriction... however, they are nonabelian as expected.

Another thing: is there any way we can generalize that? because I don't know how I may use n^2=1 (mod 2pq) for instance.

10. Anyway, I've been studying groups of order 42 for a while. The problem is that I got 7 non-isomorphic groups, although there are actually only 6. So one of them must be wrong (because all are indeed non-isomorphic, so that the additional case is not even a group).

Here they are:

1) C42
1[[1]]+1[[2]]+2[[3]]+2[[6]]+6[[7]]+6[[14]]+12[[21]]+12[[42]]

2) Direct product C7xD3
1[[1]]+3[[2]]+2[[3]]+6[[7]]+18[[14]]+12[[21]]

3) Direct product C3xD7
1[[1]]+7[[2]]+2[[3]]+14[[6]]+6[[7]]+12[[21]]

4) D21
1[[1]]+21[[2]]+2[[3]]+6[[7]]+12[[21]]

5) Direct product C2 with a semidirect product C3xC7.
i.e. C2x(C3xC7)
1[[1]]+1[[2]]+14[[3]]+6[[7]]+14[[6]]+6[[14]]

6) Semidirect product C2 with a semidirect product C3xC7.
i.e. C2x(C3xC7)
or semidirect product C3xD7
1[[1]]+7[[2]]+14[[3]]+6[[7]]+14[[6]]

7)A certain semidirect product C2 with a semidirect product C3xC7.
or a certain semidirect product C7xC6
1[[1]]+21[[2]]+14[[3]]+6[[7]]

I'm sure the first four are correct.

So the incorrect case is one of the last three.

So here are some details about them so that you may find the contradiction more easily.

5) G has 7 H3 and H6, a unique H2, H7 and H14. H21 exists, it is normal but nonabelian.

6) G has 7 H2 and H3 and H6 and unique H7, H14 and H21. Neither H14 nor H21 are abelian.

7) G has 21 H2, 7 H3 and a unique H7. Also, 7 nonabelian H6, 3 nonabelian H14 and a unique nonabelian H21.

11. Originally Posted by Stranger
Could you tell me what field of math the above stuff is dealing with?
Group theory. A very charming branch of Abstract Algebra and my favorite course.

The idea is that a field like Analysis for example always deals with distances, which makes it a bit "static" branch of pure mathematics. But Algebra always deals with operations, and that's why it's very dynamic. Your groups are alive!

But I love Analysis too. What I don't really like is things like "series solutions for differential equations", "numerical analysis"...etc all of that is quite boring for me at least
Thanks. Looks interesting, I'll have to look it up on Wikipedia. Are you taking it as a full group theory course or something?

12. Originally Posted by Chemboy

Thanks. Looks interesting, I'll have to look it up on Wikipedia. Are you taking it as a full group theory course or something?
That's my second course in Group theory and my last one as an undergraduate. There is still a lot to do in postgraduate studies.

Wikipedia is a great website, but I beleive you need a reference. I usually check wikipedia when I want to get a general idea on something, or when I'm searching about a specific thing... besides that's math, so you have to solve problems to make sure you understand and to make the thing more interesting.

So, if you can't borrow mathematics books from your library (unfortunately that's the case in my faculty, only mathematicians can borrow books, so that I always have to borrow for my Physics friends), if you can't you may check www.gigapedia.org. They have many interesting books there. I beleive Humphrey's book "A course in group theory" is the easiest book to start with.

And as serpicojr says, you'll also find other books there like "group theory in chemistry" "group theory and spectroscopy", so that this theory indeed have applications in chemistry, and of course in Physics.

Let me warn you however that the first elementary concepts will not seem very interesting... check whether that is a group, write the multiplication table for that group... I remember my impatience when I had to check whether an operation was associative. There is no getting around though, because you'll have to use these things later. Things will start to get interesting when you learn what is a subgroup, what bounds are on it... but the real fun only starts with Lagrange Theorem, one of the most important theorems in group theory.

An ideal target of group theory is the classification of all groups. That's a bit like biologists classifying animals and chemists classifying elements. This is far from being easy, and in any reference, the classifcation of the first 15 or 30 groups comes very late. There is however an infinite family you'll be able to classify after you learn some theorems. These are the "cyclic groups", the easiest groups you can deal with. Much more complicated groups were also classified: the abelian groups. But that's not exactly an easy thing.

From a certain point of view, symmetric groups are the most important family of groups. That's because you'll later understand that any group "resembles" a subgroup of a symmetric group.

13. Have you made any progress here? I haven't really thought about it, but it looks like 6 and 7 may be the same. I don't know... I'm kind of sick and out of it and am pretty useless at the moment.

14. Yeah I just found it at last.

It's the last group that's wrong. I'll have to review my calculations now to see why.

Thanks

15. By the way, I've been thinking about groups of order p^3 for a while.

I was able to prove that if G is nonabelian, then all elements of G lie in subgroups of order p^2. This is trivially true for elements of order p^2, but the idea is that even all elements of order p are also in subgroups of order p^2.

I've been trying to generalize that for any group of order p^3. Is there any theorem discussing that? maybe even for a p-group in general? (in non-cyclic cases of course)

16. Also, you say

note that an automorphism of C65 is determined by the image of a generator, say 1 (viewing C65 as Z/65Z). 1 must be sent to another generator in order for this to be an automorphism, so 1 can be sent to any integer relatively prime to 65
What if the group is not cyclic. For instance, I want an automorphism of D7.

How can I get that?

Finally, is there an easy way to calculate |Aut(G)| in some cases?

17. Your conjecture is true. Suppose G is nonabelian and has order p<sup>3</sup>. Let g be a nontrivial element of G. We consider three cases:

1. g has order p<sup>2</sup>. Then it's in the cyclic group generated by g has the desired properties.
2. g is in the center and has order p. There exists a nontrivial element h not in the center (G is nonabelian). If h is of order p, then <h,g> is a group of order p<sup>2</sup> containing g. If h is of order p<sup>2</sup>, then <h<sup>p</sup>,g> is a group of order p<sup>2</sup> containing g.
3. g is not in the center and has order p. The center is nontrivial, so take an element h in the center. Then <g,h> or <g,h<sup>p</sup>> is a subgroup with the desired properties depending as to whether h has order p or p<sup>2</sup>.
---------------------------------
Calculating the automorphism groups of nonabelian groups is more difficult. I don't know a lot about this. The group always generates a certain number of automorphisms of itself--the "inner" automorphisms. These are automorphisms which arise by conjugation, i.e.:

f<sub>g</sub>(h) = ghg<sup>-1</sup>

This is sometimes the whole story. Sometimes it isn't. The other automorphisms are "outer" automorphisms. The inner automorphisms form a normal subgroup in the automorphism group, and the outer automorphism group is the quotient of these two. So, for example, we have:

Aut(S<sub>n</sub>) = S<sub>n</sub> for n ≠ 2, 6
Aut(S<sub>2</sub>) = 1
Aut(S<sub>6</sub>) is a semidirect product of S<sub>6</sub> and C<sub>2</sub> (and we could then say Out(S<sub>6</sub>) = C<sub>2</sub>)

If you're interested in this, look up the proofs for these facts, look up the facts and proofs for alternating groups, dihedral groups, etc., and see what sorts of techniques are used.

18. nevermind what I just posted; I reviewed my calculations.

Thanks very much. I'll try to look for the automorphisms you mentioned.

19. I read what you had posted but didn't get a chance to respond. What part of what you posted did you change your mind about?

20. sorry I'm a bit confused... I guess I'm tired. No, my calculations seem to be correct after all. So what do you think? I managed to write a proof for the general statement. It's quite long so I can send it to you if you want. Unless of course you have an example to disprove the statement, in which case my proof would be useless...

21. And I'm really sick of this last case I wrote in the 42. I still can't find the contradiction and it's getting really boring for me. The reason why I'm saying it's the last case that's wrong is that article

http://www.emis.de/journals/NYJM/j/2003/9-8.pdf

That of course if I understood what he said. It's quite advanced for me. But he talks about the 2pq in his 3rd section. I just don't understand what he says... in page 106 he says something about the automorphisms which I don't understand. alpha(-1) = etc....

Apparently the key is there. In his first case he found 4 non-isomorphic groups. Something we proved for 130, but I don't understand his general argument. And in his second case he found only 2 non-isomorphic group while I found 3 (which is the problem).

Could you please explain what he's saying? My vacation is ending today, so I promise I'll stop bothering you after I get rid of this case

22. Okay, so let's pick this up where (let's assume this is a she) she says that:

G = J x<sub>α</sub> Z<sub>2</sub>

This means G is the semidirect product of J and Z<sub>2</sub> with α being a homomorphism from Z<sub>2</sub> to Aut(J), so that the group operation is given in terms of the semidirect product as(writing things multiplicatively):

(a,c)(b,d) = (a [α(c)](b),cd)

with a, b elements of J, c, d elements of Z<sub>2</sub>. Now we have a few cases for the structure of J. She's already shown that J itself is a semidirect product of Z<sub>p</sub> and Z<sub>q</sub>.

If this is indeed the direct product, then this is the same thing as J being Z<sub>pq</sub>. You can calculate that Aut(J) = Z<sub>p-1</sub>xZ<sub>q-1</sub> in this case. What are the homomorphisms from Z<sub>2</sub> to this group? Well, we just have to find the possible images of the nontrivial element of Z<sub>2</sub>, which has order 2, so we have to find the elements of order dividing 2 in Aut(Z<sub>pq</sub>). Evidently, she is making a couple of conventions here. She is viewing the group Z<sub>2</sub> as the isomorphic group {Â±1}, which is literally 1 and -1 under multiplication. -1 is the nontrivial element of this group. She's also viewing, for any prime l, the group Z<sub>l-1</sub> as the "multiplicative group of units mod l", also written as Z<sub>l</sub><sup>x</sup>. This is just the numbers 1 through l-1 considering modulo l with the group operation being multiplication. You can check that this is a group. So the elements of order dividing 2 are Â±1, and so the elements of order dividing 2 in Aut(Z<sub>pq</sub>) = Z<sub>p-1</sub>xZ<sub>q-1</sub> are (Â±1,Â±1). Some explicit computations will give you the group isomorphisms she writes down, but the basic deal is that, for any cyclic group Z<sub>n</sub>, there is an automorphism given by inverting each element. If you take the semidirect product of this group and Z<sub>2</sub> corresponding to the homomorphism sending "-1" to this automorphism, you get D<sub>n</sub>. So "different pieces of the semidirect product turn into dihedral groups depending on which pieces are inverted by -1".

The next part gets a little more involved and requires me reading the paper more. Namely, I don't know this notation "Hol". For you to get a good understanding of the proof, you'll have to read the paper yourself. I'd be happy to help you with anything you don't understand.

In any case, I believe the result in the paper. I've gotten lazy in my older years and would either: a. just be content that the result is stated here, or; b. if I really wanted to prove the facts for 42, I'd go through the paper letting p = 7, q = 3, and proving things for myself in these explicit cases.

23. That problem angered me so much I promised myself I wouldn't think about it anymore....

The thing I found that doesn't seem nice about last case is that it's the only case where the commutator subgroup is nonabelian. In all other cases the commutator subgroup is cyclic. As a consequence, the length of the derived series of the subgroups is much longer here.

Another consequence is that you can prove that the order of the product of any two elements of order 2 gives an element of order p or q. If it was p, q or pq it would have been ok (apparently this happens in Dpq if I'm correct), so I need to find 2 elements of order 2 such that their product have order pq.

The elements I suspect are apa2ap^-1 and aqa2aq^-1 where ak are elements of order k. Note that both of these have order 2 since we have pq conjugate a2.

But as I can't prove it I won't try again. That's the end of it.

24. mmm I can't give up. I have to admit I'm not strong enough to solve the problem right now... I have to learn some more.

Ok, I want to make sure some of my elementary notions are correct.

I know the union of 2 groups is not necessarily a group. However, are there certain conditions sufficient for this union to be a group?

Also, there is I beleive something called the "join" of two groups. I beleive this is <H union K> where H and K are 2 groups. Is there any way to know the order of that new group? I beleive H and K are 2 subgroups of that thing. Am I correct? And when will A union K equal that thing?

That's it for now. Thanks a lot.

25. Let G and H be groups (contained in some bigger group), and suppose the union of G and H is a group. Then I claim that one of G or H contains the other. Suppose not. Let x be an element of G\H, y an element of H\G. Then xy is an element of the union of G and H, so it must be in either G or H. If xy is in G, then so is x<sup>-1</sup>xy = y, a contradiction. Similarly, if xy is in H, then so is x, a contradiction.

The join of two groups can be a confusing thing. For example, define two matrices:

S = [[0,1][-1,0]]
R = [[0,1][-1,1]]

(where this notation means [[x,y][z,w]] is the 2x2 matrix with upper row [x,y], lower row [z,w]). Then S has order 4 and R has order 6. But the join of the cyclic subgroups generated by S and R, i.e. the group <S,R>, is SL<sub>2</sub>(Z), all 2x2 matrices with integer coefficients, an infinite group.

You may suggest that if we pick two elements from a finite group, then we can say something about the size of the group they generate, but this isn't even the case. For any integer n > 1, consider the group SL<sub>2</sub>(Z/nZ). This is the group of 2x2 matrices with entries in the ring Z/nZ (if you don't know what a ring is, then just think of this as being addition and multiplication modulo n). It turns out there is a surjective map SL<sub>2</sub>(Z) -> SL<sub>2</sub>(Z/nZ), and so the images of R and S generate the latter group. And SL<sub>2</sub>(Z/nZ) has arbitrarily large order; for example, when n = p is prime, then the order is p<sup>3</sup>-p.

----------------

Big edit: SL<sub>2</sub> also requires that the matrices have determinant 1. I can't believe I forgot that.

26. mmm... but your argument about the union is restricted to the union of 2 groups.

For instance, the union of the 3 groups {(0,0),(0,1)} union {(0,0),(1,0)} union {(0,0),(1,1)} is the Klein four group.

mmm... so the union of 2 groups is not a group unless one of them contains the other... so if none contains the other, then we are sure that <A union B> is bigger than A union B. How big is the problem... I was just going to ask whether we could find an upper bound for the order of that group. But you're saying it can be infinite...

I beleive we can say that the union of a family of groups Ai is a group if and only if <U{Ai}> = U{Ai}. Because if U{Ai} is a group, then since it is closed, it generates itself. And if <U{Ai}> = U{Ai} then U{Ai} is obviously a group... still, this identity isn't really useful...

27. I came across interesting facts about automorphisms of cyclic and dihedral groups:

From Herstein (Topics in Algebra) page 69, Aut(Zn) is isomorphic to Un, where Zn is a cyclic group of order n and Un the group of integers relatively prime to n under multiplication mod n.

Little question: isn't |Un| = phi(n), where phi is Euler Function?

Furthermore, according to a result in number theory called the Primitive Root Theorem, Un is cyclic iff n = 1, 2, 4, p^k or 2p^k.

Finally, according to a research

|Aut(Dn)| = n |Aut(Zn)|

If the answer to my little question is "yes", then we can say that

|Aut(Dn)| = n phi(n)

That's really charming!

Also, Aut(Z) is isomorhic to Z2, where Z is an infinite cyclic group. (Herstein again, same page)

28. I finally found the bug in last case!

It turns out that according to an exercise, if K is a normal cyclic subgroup of G, then the commutator subgroup of G must be a subgroup of the centralizer of K. As this is not the case here (Q is normal cyclic and equals its centralizer but G' = PQ) we have a contradiction.

edit: er sorry, it turns out I don't need the exercises I just posted (if you read them). I need another one, but I'll try to solve it first before I bother you with it.

Thanks again!

29. Ok, here is the exercise, but don't bother solving it. Apparently it relies on some other exercises and I'm really too tired to solve them all. So I'll put it here just for general knowledge

Assume p is an odd prime and G is a p-group. Then
All subgroups of G are normal ==> G is abelian

Clearly this doesn't hold for even primes (i.e. 2) since Q8 is nonabelian. As I said, it looks that thing is not easy to prove, so I'll take it as a given

Ok the last thing I'm gonna ask. Really. It's about Heisenberg group modulo p. Here it is if you want a reminder:

http://en.wikipedia.org/wiki/Heisenberg_group

Now, this is a nonabelian group of order p^3 in which every nonidentity element of G has order p.

How come then, my book wants me to prove that if p = 2, then that thing is isomorphic to D4???

Doesn't D4 contain elements of order 4? (I'm using the notation Dn not D2n).

The book is Dummit and Foote 3rd edition page 187 exercise 25 (last one)

30. Originally Posted by Stranger
Little question: isn't |Un| = phi(n), where phi is Euler Function?

Furthermore, according to a result in number theory called the Primitive Root Theorem, Un is cyclic iff n = 1, 2, 4, p^k or 2p^k.
Indeed, these are both true! I'll be getting to the former fact in my number theory topic, and perhaps I'll talk about the latter. It's a really great fact that those groups are cyclic. A particularly neat thing is, if p is prime and n is an integer that generates U<sub>p<sup>2</sup></sub>, then it generates U<sub>p<sup>k</sup></sub> for any k ≥ 1!

31. Originally Posted by Stranger
Now, this is a nonabelian group of order p^3 in which every nonidentity element of G has order p.

How come then, my book wants me to prove that if p = 2, then that thing is isomorphic to D4???
Evidently, the statement about orders doesn't hold for p = 2. Check again that your book doesn't say something like "for p ≥ 3" or, sneakier, "for odd p". Anyway, consider the matrix (praying for LaTeX soon):

111
011
001

Its square is:

101
010
001

32. mmm, so xy has order 4.

Ok thanks very much.

I guess I'll leave you alone for a while

33. Hi there, I hope you're dong well.

There is something I noticed in many groups:

Let ap1, ap2, ..., apk be elements of G of order p1, p2, ..., pk where pi are distinct primes.

Then <ap1, ap2, ..., apk> is a subgroup of G that does not contain an element of order apk+1, where apk+1 is an element of prime order pk+1, not one of the generating elements. i.e. these elements generate a subgroup of order (p1^m1)x(p2^m2)x...x(pk^mk) for some mi, but never (p1^m1)x...x(pk^mk)x(pk+1^mk+1).

Note that in my generating set, I take only one element of each order. I'm saying that because sometimes ap1 x ap1 can give ap2 where p1 and p2 are distinct primes. Also, if I'm correct, Dpq can be generated by pq conjugate elements of order 2. But this is not the case here.

In particular, if p1, p2, ..., pn are n distinct primes that divide |G|, then the only way that could generate G is <ap1, ap2, ..., apn>. I'm not saying this will generate G. I'm just saying that G will never be generated if we miss one of these api in our generating set.

I beleive this generally holds. But I don't know how to prove it. Do you have any idea? It would be great for me.

Thanks very much as always!

34. By the way, when I said ap1 x ap1 gives ap2, I didn't mean squaring ap1 of course. That would be nonsense. I meant the product of ap1 by another ap1 in a different subgroup. This seems to happen when the 2 subgroups are conjugate. This is why I'm saying I'm taking just one element of each order. Obviously, I'm also taking (p1-1) ap1, but the idea is that they all lie in the same cyclic subgroup, not in conjugate subgroups.

Anyway, I believe my conjecture is equivalent to the following conjecture:

If ap1 and ap2 are 2 elements of G of order p1 and p2 for some distinct primes p1 and p2, then |ap1ap2| = (p1)^m1x(p2)^m2 for some m1, m2 equal or greater than zero.

If they commute, then |ap1ap2|= p1p2 from a previous result. The idea is when they do not commute...

I've been thinking about the following fact:

|HK| = |H| x |K| / (H n K). This generally holds even when HK is not a subgroup. I don't know... any idea?

35. I hate to burst bubbles. Looking at the symmetric group on 5 letters, I found the following. Consider the group generates by (12) and (12345). It contains the following elements:

(12)(12345)(12) = (13452)

(12345)(13452) = (14)(35)

(13452)(12345) = (24)(35)

(14)(35)(24)(35) = (142)

And, of course, this last element has order 3. In fact, it looks like you may be able to generate all of S<sub>5</sub> using just these two elements.

36. It's ok, I got used to that

Well, the idea is that I'm trying to prove that if G is a group of order p1p2...pn for some distinct primes pi (i.e. a group of order positive square free integer), then G is not simple. That's not an exercise I encountered, it's just a feeling I have.

I proved it for pq and pqr, but I can't generalize the methods I used there. So I've been thinking that most probably, the group of order p2p3...pn is normal. Actually, it will automatically be normal if I can prove that it exists... so I thought I'd take the elements <ap2, ap3,..., apn> and say that they generate this subgroup, say P2P3...Pn. If not, then they generate G, but I've been trying to prove they cannot generate G, but in vain.

So, any idea? The first thing I tried to do is to say that one Sylow subgroup is normal (most probably the last one of order pn) but I couldn't prove it either...

Thanks a lot!

37. I don't want you to give me the proof of course... but I could use some hints or suggestions...

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