Chủ đề này có 6 trả lời

[nemo]

Suppose that G is a group satisfying the Ascending Chain Condition (ACC) and the Descending Chain Condition (DCC) on normal groups.

Show that if H,J,K are groups such that G is isomorphic to the direct product HxJ of H and J, and G is isomorphic to HxK, then J,K are isomorphic.

Since H,J,K must satisfy the ACC and DCC on normal subgroups, this exercise seems to require a straightforward application of the Krull-Schmidt Theorem, but I don't see how to do this, in other words, I don't know that theorem. Anyone have any ideas ?

[canh_dieu]

The Krull-Schmidt Theorem states that if http://dientuvietnam...n/mimetex.cgi?G is a group satisfying both the ACC and DCC conditions on normal supgroups then it decomposes into direct product of a finite number of indecomposable groups. This decomposition is unique up to isomorphism and permutation.

Knowing that http://dientuvietnam.../mimetex.cgi?G. I think now it's not difficult to conclude that http://dientuvietnam...n/mimetex.cgi?J and http://dientuvietnam...n/mimetex.cgi?K are isomorphic using the uniqueness property mentioned in the Krull-Schmidt Theorem.

For an acount of ACC and DCC conditions and the Krull-Schmidt Theorem, see Hungerford's book (Algebra, GTM).

[nemo]

For an acount of ACC and DCC conditions and the Krull-Schmidt Theorem, see Hungerford's book (Algebra, GTM).

Yes, I am going to do it soon, I just has that book. Thanks so much !

The Krull-Schmidt Theorem states that if  is a group satisfying both the ACC and DCC conditions on normal supgroups then it decomposes into direct product of a finite number of indecomposable groups. This decomposition is unique up to isomorphism and permutation

Thus, Krull-Schmidt is the main idea.

Part of the problem though is that http://dientuvietnam...mimetex.cgi?HxJ need not be equal to http://dientuvietnam...imetex.cgi?HxK; they simply need to be isomorphic. If we let H have indecomposable factors http://dientuvietnam...gi?H_1,...,H_m. J have indecomposable factors http://dientuvietnam...cgi?J_1,...,J_n, and K have factors http://dientuvietnam...cgi?K_1,...,K_p, then we can conclude that p=n. Let http://dientuvietnam...mimetex.cgi?i_1 be injection of H into http://dientuvietnam...mimetex.cgi?HxK and http://dientuvietnam...mimetex.cgi?i_2 be injection of K into http://dientuvietnam...imetex.cgi?HxK. Let f be the isomorphism from http://dientuvietnam...mimetex.cgi?HxK to http://dientuvietnam.net/cgi-bin/mimetex.cgi?HxJ. Then http://dientuvietnam.net/cgi-bin/mimetex.cgi?G=HxJ=f(HxK)=f(i_1(H))xf(i_2(K)). One can conclude that http://dientuvietnam.net/cgi-bin/mimetex.cgi?H_r and http://dientuvietnam.net/cgi-bin/mimetex.cgi?f(i_1(H_r)) are isomorphic for , however, it is not obvious to me how this helps my cause. Any thoughts?

p/s: What is GTM ?

[canh_dieu]

GTM = Graduate Texts in Mathematics

A\times B

http://dientuvietnam.net/cgi-bin/mimetex.cgi?m.

[nemo]

I think you can proceed by induction on .

Yes, I see.

[vinhspiderman]

Remark: This Krull - Schmidt theorem for groups is likely to the Jordan - Holder theorem, which is for finite lenth modules:

If a module M is finite lendth then every "composition series" of M has the same lenth and the elements in the composition series are some permutation of the elements in an arbitrary choosen composition series.

[nemo]

To have a connection with the Jordan - Holder theorem, I need a reference for Jordan-Holder Theorem for topological groups. I tried to prove it by imitating the proof of Jordan-Holder Theorem for R-Modules but I found I should assume that the locally and segma-compact to use the third isomorphism theorem which is only hold under this condition for the topological groups.