Two more simple cyclic group problems

[radou]

Hm, km = jm (mod n) means that n divides km - jm = m(k - j). Now, I'm not sure about this, I tried it out on some numeric examples and it seems to work, and since you're here to ask, does it follow that n divides k - j, since m and n are relatively prime (i.e. m cannot divide n)?

P.S. If this was a stupid question, forgive me, I'm still warming up on this topic.

 

[micromass]

Yes, you are correct: if n divides m(k-j), then it must imply that n divides k-j (if n and m are coprime). But you have to prove it rigourously...

 

[radou]

Yes, I know, I'll try to prove it. Once I do so, the rest is trivial, since then a^k = a^(qn+j) = a^(qn)a^j = a^j.

 

[micromass]

What does Bezout's theorem give you in modulo n?

 

[radou]

Hm, no hints yet, but can this be proven so that I assume the contrary, i.e. that n doesn't divide k - j either? I'm messing with some inequalities and trying to apply Bezout's theorem, but I'm still getting nowhere. If the proof involvec modulo n congruence classes, I'm still not into that yet, so there may be something I don't see right away...

 

[micromass]

I'm not sure, I would go for a direct proof. I think that is easiest...
But I suppose contradiction may work, but I don' know how at this point...

The proof is easiest if you work with modulo congruence classes. But I understand that you don't really want to use it because you don't feel comfortable with it. But after you figured out the proof for this, I really suggest transforming the proof so that it works with modulo n to. Modulo n congruences is really a very important tool algebra!

 

[radou]

I'm not sure, I would go for a direct proof. I think that is easiest...
But I suppose contradiction may work, but I don' know how at this point...

The proof is easiest if you work with modulo congruence classes. But I understand that you don't really want to use it because you don't feel comfortable with it. But after you figured out the proof for this, I really suggest transforming the proof so that it works with modulo n to. Modulo n congruences is really a very important tool algebra!

Hmm, OK, I'll skip to Chapter 23 then for a while, I hope it will help to solve this problem finally...

 

[radou]

Btw, I don't understand why the author put this exercise in a section too early... i.e. if there is no other "decent" way to prove this without using congruence modulo classes... Specially considering the fact that most of the exercises are far more easy than this one.

 

[micromass]

Well, you don't need to use module classes. The proof is not that difficult if you don't use them. However, with module classes, the proof is more natural. That's why I prefer it that way.

It's of course another possibility that there's a really elegant proof for this, but I really don't see it. What we've done so far is the way I've seen it, and how it is in all the books I've seen. However, the fact that this exercise comes so early, might indicate that there is a short proof of this...

So in short, you don't need chapter 22 for this, but I certainly suggest coming back to the problem once you did chapter 22. Since the proof will become more natural and easier then what it is now...

 

[radou]

Hmm, yes...

I started with m(k - j) = pn. I tried to assume n doesn't divide k - j, so we have k - j$$\neq$$ zn, for every integer z. I tried to multiply with m and apply Bezout's theorem in order to obtain a contradiction somehow, but I think this won't work. Any thoughts perhaps?

I hate to say that, but it's probably the best to leave that exercise for later.

 

[micromass]

Alright, I'll give a hint. It should be easy once you did this:

Bezout's theorem gives us an x and y such that mx+yn=1.

Now, assume that m(k-j)=pn. Now multiply both sides by x. This should solve the problem. Note that it isn't clear why multiplying by x gives a nice answer, but this will become clear once you studied modulo congruences...

 

[radou]

Alright, I'll give a hint. It should be easy once you did this:

Bezout's theorem gives us an x and y such that mx+yn=1.

Now, assume that m(k-j)=pn. Now multiply both sides by x. This should solve the problem. Note that it isn't clear why multiplying by x gives a nice answer, but this will become clear once you studied modulo congruences...

Hm, I can't see how this gives the solution... And how do we know that x is non-zero?

 

[micromass]

Well, if x is zero, then yn=1, thus n=1. Then we are working in the cyclic group with 1 element (i.e. the trivial group).

If x is nonzero, then we multiply both sides of m(k-j)=pn by x. We get xm(k-j)=pn. But since xm+yn=1, this gives us (1-yn)(k-j)=pn. Working out the brackets gives us k-j=n(p+yk-jy).
Thus we have found a natural number q such that k-j=qn. This yields that a^k=a^j.

 

[radou]

Well, if x is zero, then yn=1, thus n=1. Then we are working in the cyclic group with 1 element (i.e. the trivial group).

If x is nonzero, then we multiply both sides of m(k-j)=pn by x. We get xm(k-j)=pn. But since xm+yn=1, this gives us (1-yn)(k-j)=pn. Working out the brackets gives us k-j=n(p+yk-jy).
Thus we have found a natural number q such that k-j=qn. This yields that a^k=a^j.

Hm, after multiplying both sides of m(k - j) = pn by x, don't we get xm(k - j) = xpn? But that doesn't matter, since (1 - yn)(k - j) = xpn ==> k - j = yn(k - j) + xpn = (y(k - j) + xp)n, which demonstrates that our mapping is injective.

 

[micromass]

Hm, after multiplying both sides of m(k - j) = pn by x, don't we get xm(k - j) = xpn?

Yes, I'm sorry. Of course I should have multiplied both sides... But it doesn't matter that much...

 

[radou]

OK, no problem! Then there's only surjectivity left to show, i.e. let b = a^k be an element of <a>, we need to show that there exists some integer j such that (a^j)^m = a^k. Now, I used an exercise from the next exercise section to prove this (which probably implies there's another way to prove it): since m and n are relatively prime, a^m is a generator of <a>. Then clearly there exists an integer j such that (a^m)^j = (a^j)^m = a^k.

 

[micromass]

Yes, this is correct. But note that there are some more ways of proving this.

The easiest way is by noting the following: if X and Y are finite sets such that |X|=|Y| (i.e. X and Y have the same number of elements) and if f f:X--> Y is a function, then f is injective iff it is surjective.

This immediately implies surjectivity, since you have already proven injectivity.
It's really a handy property: if you have two finite groups G and H and you want to prove that they are isomorphic, then you just need to check that G and H have the same elements and that there exists an injective homomorphism between them. This is fun because you never need to show surjectivity...

Another way is through Bezout's theorem. Take a^k. You need to find a j such that a^(jm)=a^k. Thus you need to find a j such that jm-k=pn for some p. By Bezout's theorem, you can find xm+yn=1. Take j=xk. Then jm-k=xmk-k=k-ynk-k=ykn. Thus p=yk in this case...

But I think your proof is the cleanest possible proof for this...

 

[radou]

Yes, this is correct. But note that there are some more ways of proving this.

The easiest way is by noting the following: if X and Y are finite sets such that |X|=|Y| (i.e. X and Y have the same number of elements) and if f f:X--> Y is a function, then f is injective iff it is surjective.

Ahh, of course, I should have thought of this, since this was even an exercise in an earlier section!

Thanks!