Proof of Infinite Cyclic Group Isomorphism to Z

[radou]

I came across a theorem the proof of which I don't quite understand.

The theorem states that every infinite cyclic group is isomorphic to the additive group Z.

So, the mapping f : Z --> G given with k |--> a^k, where G = <a> is a cyclic group, is an epimorphism, which is quite obvious. Now, in order to show that it's an isomorphism, one should show that it's a monomorphism, i.e. that the kernel is trivial. So, we have Ker f = {k from Z such that a^k = e}. Now, Ker f contains 0, but how do I show that it only contains 0?

 

[StatusX]

Use the fact that the groups infinite.

 

[radou]

Use the fact that the groups infinite.

OK, but I'm not really sure how. The book says that, since Ker f is a subgroup of Z, and if it's non trivial, then we have Ker f = <m>, where m is the least positive integer such that a^m = e. I understand that, but I don't see how it proves anything.

 

[matt grime]

The group is cyclic, so it has a generator, x, so every element is equal to x^n for some integer n. x has infinite order, since the group is infinite, hence x^n=x^m if and only if n=m. Hence the group has an obvious isomorphism with Z.

 

[Office_Shredder]

Suppose a^k=e, k=/= 0. Then a^k+1 = a, and you just discovered your group is finite

 

[radou]

hence x^n=x^m if and only if n=m

I'm probably missing something, but where does this fact about injectivity come from?

I know the isomorphism is obvious, but I don't understand some formal details.

 

[mathwonk]

you have proved every cyclic groupis isomorphic to a quotient of Z. the only quotient that is infinite is the quotienht by {0}.

 

[matt grime]

I'm probably missing something, but where does this fact about injectivity come from?

I know the isomorphism is obvious, but I don't understand some formal details.

Suppose not, then... (this is something you must have proved on the first exercise sheet in the course)

 

[radou]

Okay, so suppose there exist integers p and q, p=/=q, such that a^p = a^q. Then we have a^(p-q) = e, so p - q belongs to Ker f = <m>, where m is the least positive integer such that a^m = e.

Again, I'm not sure how to proceed.

 

[matt grime]

No, you're doing this the wrong way. I said x (which you relabelled a) was a generator of this notional infinite cyclic group. Obivously x^m is never e for a generator of an infinite cyclic group unless m=0. Look, the infinite cyclic group is {x^n : n in Z}, so the isomorphism sends x^n to n. It is trivially an isomorphism, the question really has no content: it is just taking logs base x.

 

[radou]

No, you're doing this the wrong way. I said x (which you relabelled a) was a generator of this notional infinite cyclic group. Obivously x^m is never e for a generator of an infinite cyclic group unless m=0. Look, the infinite cyclic group is {x^n : n in Z}, so the isomorphism sends x^n to n. It is trivially an isomorphism, the question really has no content: it is just taking logs base x.

OK, I understand now. I shouldn't have wasted so much time on this.

 

[radou]

Btw, I should have realized earlier how trivial it can be proved.

Assume Ker f =/= {0}. Ker f = <m>, where m is the least positive integer such that a^m = e. Hence, a^(m+1) = a^m*a = e*a = a , a^(m+2) = a^(m+1)*a = a^2 , ... , a^(2m) = a^m*a^m = e, etc. Therefore we have a contradiction with the fact that G is an infinite cyclic group, so Ker f = {0}, which makes f a monomorphism, and hence an isomorphism.

 

[radou]

***

I don't want to start a new thread, so I'll simply post another thing I'm not sure about in this thread, hoping it will be noticed.

I have to prove that, an infinite group is cyclic iff it is isomorphic to each of its proper subgroups.

(=>) OK, let G = <a> = {a^n : n e Z} be an infinite cyclic group. Since G is cyclic, so is every of its subgroups. Let H < G, where H is a proper subgroup. Then H =<a^m> = {(a^m)^n : n e Z}, where m is the least positive integer such that a^m e H. Consider the mapping f: G --> H given with f : a^n --> (a^m)^n. It is a homomorphism, since f(a^p*a^q)=f(a^(p+q))= (a^m)^(p+q) = (a^m)^p*(a^m)^q = f(a^p)*f(a^q). Further on, it is a monomorphism, since (a^m)^p = (a^m)^q => p = q, since a^m is in G and the elements of G are all distinct. Clearly it is an epimorphism, since I am f = {(a^m)^n : n e Z} (since G is isomorphic to Z), which makes it an isomoprhism, so G is isomorphic to H.

(<=) This is the direction I'm having problems with, so I'd be grateful if someone checked my work and helped me out with a further hint/correction.

 

[StatusX]

For the first direction, have you proved every subgroup of a cyclic group is cyclic? This is the main part of the proof, but you seemed to have skipped this and focused on the more trivial details. Once you get this, you can just use the fact you showed above that all infinite cyclic groups are isomorphic to Z, and so to each other.

For the other direction, find an x in G that isn't a generator, and use the hypothesis that <x> is isomorphic to G.

 

[radou]

For the first direction, have you proved every subgroup of a cyclic group is cyclic? This is the main part of the proof, but you seemed to have skipped this and focused on the more trivial details.

It is a proved theorem in my book, so I used it.

For the other direction, find an x in G that isn't a generator, and use the hypothesis that <x> is isomorphic to G.

OK, so let the infinite group G be isomorphic to each of its proper subgroups. Let H = <a> be a nontrivial cyclic subgroup of G. But, how do I know a isn't a generator of G? (I can only assume, but that's not enough, I suppose.)

 

[StatusX]

Let H be a non-trivial proper subgroup of G, and pick an element from H.

 

[radou]

Let a e H, where H < G. Since G is isomorphic to H, for every a^n where n is an integer, there exists a unique element from G mapped to that element, and since G is infinite H = <a> is an infinite cyclic subgroup, and hence isomorphic to Z, which makes G isomorphic to Z, so G is an infinite cyclic group. (I feel there is something wrong with my logic somewhere.)

 

[StatusX]

Since G is isomorphic to H,

This is true, but irrelevant.

for every a^n where n is an integer, there exists a unique element from G mapped to that element,

What does this have to do with anything?

and since G is infinite

How do you know this?

H = <a> is an infinite cyclic subgroup, and hence isomorphic to Z, which makes G isomorphic to Z, so G is an infinite cyclic group.

H is not necessarily <a>. Also, there's an unnecesary excursion here to Z and then back again.

I think you're overthinking this, it's a one or two-liner.

 

[radou]

Well, it is known that G is an infinite group (see my first post on this problem).

If I pick a e H, where H < G, I can generate another subgroup <a> of H, right? Now, I don't know anything about it, only that it is isomorphic to the infinite group G. If what I wrote above doesn't make any sense, then how should I reason? Does, in general, an isomorphism between an infinite group (G in our case) and another group, necessarily make that group infinite, too?

 

[StatusX]

I missed that line about G being infinite. I thought you wanted to show something like this:

"Show that a group G is infinite cyclic iff it is isomorphic to all of its non-trivial proper subgroups (and such that there's at least one such subgroup, to exclude groups of prime order)."

Then the proof is pretty much the same, except you have the extra step that if G is isomorphic to a proper subgroup, then as a set it has the same cardinality as a proper subset, and so both must be infinite.

And yes, an isomorphism between groups means they differ essentially only in notation (although the exact nature of the isomorphism is often important, or else automorphism groups wouldn't contain any information). So one is cyclic iff the other is, one is infinite iff the other is, etc.

 

[radou]

OK, thanks for your help!

 

[radou]

***

Well, another little problem.

If G is a cyclic group of order n, and if k|n, then G has exactly one subgroup of order k.

Since the order of G is n, n is the least positive integer such that a^n = e. Because k divides n, there exists an integer p such that n = pk. So, we have a^n = a^(pk) = (a^p)^k = e. Now, I know that if I knew the order of the cyclic subgroup <a^p> was k, then (a^p)^k = e would hold. But I don't know if the reverse holds. If it would hold, the proof is done, since the uniqueness of the integer p implies the uniqueness of the subgroup <a^p> of order k.

Please help me with the probable flaws in my reasoning. ()

 

[StatusX]

Then the order of an element b is by definition the smallest positive integer m such b^m=e. From this you can show that if the order of a is n and n=pq, then the order of a^p is q. I don't understand your last sentence about uniqueness though.

 

[radou]

If (a^p)^q = e, how do I show that q is the least positive integer which generates the identity? If I assume there exists an integer r < p such that (a^p)^r = e, I don't have any further ideas.

 

[StatusX]

Really? Think about what's going on. If you find a lower power of a^p than the qth that's e, how does this give you a lower power of a than the pqth that's e, which would be a contradiction?

 

[radou]

Really? Think about what's going on. If you find a lower power of a^p than the qth that's e, how does this give you a lower power of a than the pqth that's e, which would be a contradiction?

Uhh, I see now, it's more than trivial.

Of course, pq = n, where n is the lowest power, so there can't exist some lower power of the qth that's e.

 

[radou]

***

Here's another one.

Let G be an abelian group, and let a, b e G such that |a| = m and |b| = n. One has to prove there exists some c e G such that its order equals the least common multiple of m and n.

Well, the least common multiple of m and n equals mn/(m, n), where (m, n) is the greatest common divisor of m and n. Since none of m, n equals 0, (m, n) exists. The book offers a hint that the case when (m, n) = 1 should be considered. So, the least common multiple of m and n is mn. So I should show that there exists c e G such that c^(mn) = e, where mn is the least positive integer which generates e. This is where I'm stuck. Any help is highly appreciated.

 

[matt grime]

What is the only element of G you can write down that might have a chance of being an element of order lcm(m,n)?

 

[radou]

What is the only element of G you can write down that might have a chance of being an element of order lcm(m,n)?

I don't think I'm following. Btw, are we assuming (m, n) = 1 or should I just forget about the hint from the book?

 

[matt grime]

You don't follow? Look, forget the hint. You're given elements a and b. The *only* thing you can do is to write down some more elements in G (e, a^-1. b^-1 and what else, perhaps?) and see what their orders are. It is easy to see, if (m,n)=1, what the order of *the only other element* you should write down is, and thence what the order is in general, which is just as easy, but a little messier.

Once more, radou, the answer is: get your hands dirty and calculate something. You've been given a and b, so you're going to have to write this element of order lcm(m,n) in terms of them.

 

[radou]

So, if I take ab, <ab> = {(ab)^p : p e Z} = {a^p*b^p : p e Z}, since G is abelian. For p = mn, (ab)^mn = e. But I'm not sure if this shows anything, since if the order q of an element c is known, then c^q = e.

 

[matt grime]

What? You know that (ab)^mn=e, so you know that the order of ab divides mn (eg lcm(m,n) is such a number...)

 

[radou]

What? You know that (ab)^mn=e, so you know that the order of ab divides mn (eg lcm(m,n) is such a number...)

I get it, I should have had the proposition which states that fact in mind. Thanks.

 

[radou]

***

Another question which confuses me.

Let H, K, N be subgroups of G such that H < K, H\cap N = K\cap N and HN = KN. Prove that H = K.

Well, the H\cap N = K\cap N part confuses ne, since, by definition,
H\cap N = \left\{x \in G : x \in H \wedge x \in N\right\}, and K\cap N = \left\{x \in G : x \in K \wedge x \in N\right\}. Since H is a subgroup of K, it is also a subset of K, and hence H \cap N must be a subset of K \cap N. Since they're equal, there doesn't exist an x from K which isn't in H, and hence H = K.

There is probably something terribly wrong with my reasoning, but unfortunately, I can't figure out what it is. Thanks in advance.

 

[matt grime]

Since, at no point, did you invoke the fact that HN=KN, you might want to rethink what that means.

 

[matt grime]

hence H \cap N must be a subset of H \cap N.

I presume you didn't mean to write that.

 

[morphism]

What are |HK| and |HN|?

 

[radou]

I presume you didn't mean to write that.

A typo - I corrected it.

Since, at no point, did you invoke the fact that HN=KN, you might want to rethink what that means.

If HN = KN, we know HN\subseteq KN, which is quite obvious, since H < K, and we know KN\subseteq HN, so x = kn e KN implies x e HN, so every k must be from H, and hence K\subseteq H, which implies K = H. But why are two facts given in the problem (HK = KN and the first one about intersections), since they lead to the same conclusion? Again, obviously my reasoning is wrong.

 

[morphism]

If HN = KN, we know HN\subseteq KN, which is quite obvious, since H < K, and we know KN\subseteq HN, so x = kn e KN implies x e HN, so every k must be from H, and hence K\subseteq H, which implies K = H. But why are two facts given in the problem (HK = KN and the first one about intersections), since they lead to the same conclusion? Again, obviously my reasoning is wrong.

Why did you ignore my reply? I'll say it again: What are |HK| and |HN|?

And your reasoning failed because kn being in HN doesn't imply k must be in H. kn in HN means there exists an h in H and an n' in N such that kn = hn', so k = hn'n^-1, which doesn't necessarily lie in H.

 

[radou]

And your reasoning failed because kn being in HN doesn't imply k must be in H. kn in HN means there exists an h in H and an n' in N such that kn = hn', so k = hn'n^-1, which doesn't necessarily lie in H.

Thanks, I got it now.

Why did you ignore my reply? I'll say it again: What are |HK| and |HN|?

Did you mean: "What are |HN| and |KN|?" Since then, |HN| = |H||N|/(H\capN) and |KN| = |K||N|/(K\capN), and, since HN = KN, and since H\capN = K\capN holds, it follows that |H| = |K|, and since H < K, we have H = K.

Edit: although, the theorem about the cardinality of HK applies only for finite subgroups, at least that's what my book says, and the problem doesn't mention they're finite.

 

[morphism]

Whoops, yes that's what I meant! And yeah, I seem to have also missed that they aren't necessarily finite.

Anyway, let k be in K. Then if kn₁ is in KN=HN, there exists an h in H and an n₂ in N such that kn₁ = hn₂. We can re-write this as h^-1k = n₂n₁^-1. The left side is in N, and the right side is in K, because H <= K. Thus h^-1k is in K\capN = H\capN.

Can you take it from here?

 

[radou]

If h^{-1} k is in H\capN, I assume it doesn't need to be true that h^-1 and k are in H\capN, too? If so, then I don't think I have any bright ideas. (I have to show that k is in H, right? It's the only inclusion we need, since H < K implies H is a subset of K.)

 

[morphism]

How is H\capN related to H?

 

[radou]

How is H\capN related to H?

It is a subgroup of H.

 

[morphism]

Right. h^-1k is in a subgroup of H (and therefore in H). So...

 

[radou]

Right. h^-1k is in a subgroup of H (and therefore in H). So...

...so h^{-1}k = h&#039;, for some h' in H\capN. After multiplying with h from the left, we have k = hh' , and hence k is in H, which proves the point. (I hope..)

 

[morphism]

Yup.(message too short)

 

[radou]

Yup.


(message too short)

Thanks a lot!

 

[radou]

***

The problem states that f : G --> H is a group homomorphism, H is abelian, and N < G, whith Ker(f) contained in N. One has to prove that N is normal in G.

The first thing that crossed my mind was a neat corrolary which stated that there is a bijective correspondence between the set of all subgroups of G containing Ker(f) and all subgroups of H, such that normal subgroups correspond to normal ones. In that case, the proof would be almost trivial, but then I remembered that the corollary required f to be an epimorphism.

Well, I know that, since H is abelian, every subgroup of H is abelian, and I know that Ker(f) < N < G, but I can't come up with anything constructive. So, any pushes in the right direction are welcome.

 

[matt grime]

Any map is surjective onto its image.