Is being closed a sufficient condition for a finite subset to be a subgroup?

[Genady]

H is a finite subset of a group G, and for each h₁, h₂ in H their combination h₁h₂ is also in H. Is H a group?

[KJW]

No. The subset also must contain the identity as well as the inverses for all elements of the subset. The associative property is however inherited from the group G.

 

Edited December 26, 2024 by KJW

[Genady]

I think, that being finite and closed guarantees that it contains the identity and all the inverses. 

Edited December 26, 2024 by Genady

[KJW]

30 minutes ago, Genady said:

I think, that being finite and closed guarantees that it contains the identity and all the inverses. 

Can you provide a proof? I can't at this time provide a counterexample.

 

EDIT: I think you may be correct. If one chooses a single non-identity element as a generator, then if it is a finite generator, the subset generated must contain the identity element and all the inverses of the elements of the generated subset, and hence be a group. But if the generator is not finite, then the subset generated is not necessarily a group.

 

Edited December 26, 2024 by KJW

[Genady]

IOW,

Let's take any h in H. Because of H being closed, all combinations h, hh, hhh, ... are in H. But because of H being finite, some combinations should repeat, say h^m = hⁿ for some m<n. Then hⁿ = h^mh^n-m = h^m and thus h^n-m is identity, e (from G). e = h^n-m is thus contained in H.

Now, if n-m = 1, then e = h is inverse of itself. Otherwise, e = hh^n-m-1, which makes h^n-m-1 an inverse of h.