Proof of Infinite Cyclic Group Isomorphism to Z

[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$\cap$N) and |KN| = |K||N|/(K$\cap$N), and, since HN = KN, and since H$\cap$N = K$\cap$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.

 

[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$\cap$N = H$\cap$N.

Can you take it from here?

 

[radou]

If $h^{-1} k$ is in H$\cap$N, I assume it doesn't need to be true that h^-1 and k are in H$\cap$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.)

 

[morphism]

How is H$\cap$N related to H?

 

[radou]

How is H$\cap$N 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 $h^{-1}k = h'$, for some h' in H$\cap$N. 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.

 

[radou]

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?

 

[matt grime]

Forget H. Replace H with Im(f).

 

[radou]

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.

 

[fawad]

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.