Proof of Infinite Cyclic Group Isomorphism to Z

In summary, the theorem states that every infinite cyclic group is isomorphic to the additive group Z. The proof involves showing that the mapping from Z to G given by k |--> a^k, where G = <a> is a cyclic group, is an epimorphism and then proving that it is also a monomorphism, leading to the conclusion that it is an isomorphism. The converse is also true, as every infinite group that is isomorphic to each of its proper subgroups must be cyclic. This can be shown by finding an element x that is not a generator in G and using the fact that <x> is isomorphic to G.
  • #36
radou said:
hence [itex]H \cap N[/itex] must be a subset of [itex]H \cap N[/itex].

I presume you didn't mean to write that.
 
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  • #37
What are |HK| and |HN|?
 
  • #38
matt grime said:
I presume you didn't mean to write that.

A typo - I corrected it.

matt grime said:
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 [itex]HN\subseteq KN[/itex], which is quite obvious, since H < K, and we know [itex]KN\subseteq HN[/itex], so x = kn e KN implies x e HN, so every k must be from H, and hence [itex]K\subseteq H[/itex], 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.
 
  • #39
radou said:
If HN = KN, we know [itex]HN\subseteq KN[/itex], which is quite obvious, since H < K, and we know [itex]KN\subseteq HN[/itex], so x = kn e KN implies x e HN, so every k must be from H, and hence [itex]K\subseteq H[/itex], 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.
 
  • #40
morphism said:
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.

morphism said:
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[itex]\cap[/itex]N) and |KN| = |K||N|/(K[itex]\cap[/itex]N), and, since HN = KN, and since H[itex]\cap[/itex]N = K[itex]\cap[/itex]N 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. :biggrin:
 
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  • #41
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[itex]\cap[/itex]N = H[itex]\cap[/itex]N.

Can you take it from here?
 
  • #42
If [itex]h^{-1} k[/itex] is in H[itex]\cap[/itex]N, I assume it doesn't need to be true that h^-1 and k are in H[itex]\cap[/itex]N, 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.)
 
  • #43
How is H[itex]\cap[/itex]N related to H?
 
  • #44
morphism said:
How is H[itex]\cap[/itex]N related to H?

It is a subgroup of H.
 
  • #45
Right. h-1k is in a subgroup of H (and therefore in H). So...
 
  • #46
morphism said:
Right. h-1k is in a subgroup of H (and therefore in H). So...

...so [itex]h^{-1}k = h'[/itex], for some h' in H[itex]\cap[/itex]N. After multiplying with h from the left, we have k = hh' , and hence k is in H, which proves the point. (I hope..)
 
  • #47
Yup.(message too short)
 
  • #48
morphism said:
Yup.


(message too short)

Thanks a lot!
 
  • #49
***

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. :rolleyes:

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.
 
  • #50
Any map is surjective onto its image.
 
  • #51
matt grime said:
Any map is surjective onto its image.

So if I look at the epimorphism f : G --> Im(f), I could apply the corollary? eg, since Im(f) and every other subgroup of H is normal, there must exist a normal subgroup of H to which N is mapped?
 
  • #52
Forget H. Replace H with Im(f).
 
  • #53
OK, so f : G --> Im(f) is an epimorphism. Since there exists a bijection between the set of all subgroups of G which contain Ker(f) and the set of all subgroups of Im(f) (where normal subgroups correspond to normal ones), the group N < G is mapped to some subgroup K < Im(f). Since K is normal (because it is a subgroup of an abelian group), N must be normal in G.
 
  • #54
can some body help me in solving the following problem.
a finite group with an even number of elements contains an even number of elements x such that x^-1 = x.
 

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