Prev: Mathematically-Rigorous Path Integration??
Next: proof of Goldbach's Conjecture
From: William Elliot on 22 Dec 2009 04:33
I'm trying to remember a simple problem.

Let H,K be subgroups of a group G.

What premise upon H and K yields the conclusion
H subgroup K or K subgroup H ?

I don't recall if G is Abelian or not.

----
From: José Carlos Santos on 22 Dec 2009 04:47
On 22-12-2009 9:33, William Elliot wrote:

> I'm trying to remember a simple problem.
>
> Let H,K be subgroups of a group G.
>
> What premise upon H and K yields the conclusion
> H subgroup K or K subgroup H ?

What about "{H,K} = {{e},G}"? It surely works.

> I don't recall if G is Abelian or not.

And I don't even understand what you mean by this.

Best regards,

Jose Carlos Santos
From: Tonico on 22 Dec 2009 06:06
On Dec 22, 11:33 am, William Elliot <ma...(a)rdrop.remove.com> wrote:
> I'm trying to remember a simple problem.
>
> Let H,K be subgroups of a group G.
>
> What premise upon H and K yields the conclusion
>         H subgroup K or K subgroup H ?



That the union H \/ K is a subgroup of G: this happens precisely iff H
sbgp. of K or K sbgp. of H.

Tonio


>
> I don't recall if G is Abelian or not.
>
> ----

From: William Elliot on 22 Dec 2009 07:15
On Tue, 22 Dec 2009, Tonico wrote:
> On Dec 22, 11:33�am, William Elliot wrote
>> I'm trying to remember a simple problem.
>>
>> Let H,K be subgroups of a group G.
>>
>> What premise upon H and K yields the conclusion
>> . . H subgroup K or K subgroup H ?
>
> That the union H \/ K is a subgroup of G: this happens precisely iff H
> sbgp. of K or K sbgp. of H.
>
Yes, that's it.
H,K, H \/ K subgroup G ==> H subgroup K or K subgroup H.

Otherwise: some h in H\K, k in K\H; hk in H \/ K.
If hk in H, then h^-1 hk = k in H, a contradiction.
Similar if hk in K.

Here's a related problem.
If H,K are convex subsets of a totally ordered group G,
.. . then K subgroup K or K subgroup H.

In view of this next problem, the two
problems are more related than I thought,

If H,K are convex subsets of a totally ordered group G,
.. . then H \/ K is a subgroup.

Proof fragments.
Assume h in H, k in K. Is hk in H \/ K? Yes.
Case 0 <= |h| <= |k|: |h| in K; hk in K
Case 0 <= |k| <= |h|: similar.

The second proposition is proven much the same as the third
or is a direct result of the first and third propositions.

Do you see any variations upon this theme or generalizations thereof?

----
From: Arturo Magidin on 22 Dec 2009 17:56
On Dec 22, 3:33 am, William Elliot <ma...(a)rdrop.remove.com> wrote:
> I'm trying to remember a simple problem.
>
> Let H,K be subgroups of a group G.
>
> What premise upon H and K yields the conclusion
>         H subgroup K or K subgroup H ?
>
> I don't recall if G is Abelian or not.

You mean, the subgroups form a chain?

Suppose G satisfies the condition, and let x and y be elements of G.
Then either <x> is in <y> or <y> is in <x>; in particular, either x is
a power of y, or y is a power of x. Thus, G must be abelian.

Moreover, G must be locally cyclic. This is possibly also sufficient.

--
Arturo Magidin
 | 
Pages: 1
Prev: Mathematically-Rigorous Path Integration??
Next: proof of Goldbach's Conjecture