# Thread: Closed subset of a group

1. ## Closed subset of a group

The problem is this:
[quote]Prove that a non-empty finite subset S of a group G is a subgroup iff it is closed under the product in G.[/tex]

It's easy to show that if the S is a subgroup then S is closed. I am having trouble with proving that S is a subgroup if it is closed. I can easily demonstrate that this is not true for an infinite subset S, but my efforts to prove the theorem for a finite subset is running into problems. I can show, for specific examples, that if the theorem is not true then either there is more than one identity in S or that S is not associative. But I'm wasting paper on the constructed examples and I really have not been able to make headway for the general case.

For right now my question is how do I prove that S contains the group identity? (I'll tackle the existence of inverses in S on my own for now. That's what I'm planning to attack until I can understand the identity part.)

-Dan

2. [QUOTE=topsquark;634231]The problem is this:
Prove that a non-empty finite subset S of a group G is a subgroup iff it is closed under the product in G.[/tex]

It's easy to show that if the S is a subgroup then S is closed. I am having trouble with proving that S is a subgroup if it is closed. I can easily demonstrate that this is not true for an infinite subset S, but my efforts to prove the theorem for a finite subset is running into problems. I can show, for specific examples, that if the theorem is not true then either there is more than one identity in S or that S is not associative. But I'm wasting paper on the constructed examples and I really have not been able to make headway for the general case.

For right now my question is how do I prove that S contains the group identity? (I'll tackle the existence of inverses in S on my own for now. That's what I'm planning to attack until I can understand the identity part.)

-Dan

Let $S:=\{s_1,...,s_n\}$ be the finite closed subset of G, and define $f: S\rightarrow S\,,\,\,f(s):=s_1s$ = multiplication

by $s_1$ (you can in fact choose any element in S that you want). Prove

1) f is 1-1 and thus onto

2) Prove now that S contains the group's unit and inverse of any element in it

3) Go celebrate since you're done.

Tonio

3. as an aside, for S finite, showing closure --> inverses is all you have to do, since if a^-1 is in S for every a, then by closure aa^-1 = e is in S as well. however, tonio's method actually requires you to show that S contains e first (this only uses the surjectivity of f), then show for each sj in S sj^-1 is in S (you need bijectivity for this).

4. Originally Posted by tonio
Let $S:=\{s_1,...,s_n\}$ be the finite closed subset of G, and define $f: S\rightarrow S\,,\,\,f(s):=s_1s$ = multiplication

by $s_1$ (you can in fact choose any element in S that you want). Prove

1) f is 1-1 and thus onto

2) Prove now that S contains the group's unit and inverse of any element in it
This is the problem. I know for a fact that the set $\{ s_1s | s \in S, s_1 \in S~\text{for fixed } s_1 \}$ contains the identity, but how do I prove that? There is a theorem related to this that I am trying to remember from my Physics version of a Group Theory text, but I can't remember the name, nor find it in the book.

Originally Posted by Deveno
as an aside, for S finite, showing closure --> inverses is all you have to do, since if a^-1 is in S for every a, then by closure aa^-1 = e is in S as well. however, tonio's method actually requires you to show that S contains e first (this only uses the surjectivity of f), then show for each sj in S sj^-1 is in S (you need bijectivity for this).
The trouble here, again, is proving that $a^{-1}$ is in S.

Once I know that the group identity e is in S I think I can do the following:

Let $a,~b \in G$ and let e be the identity in S. The solution of the following equation $ab = e$ can be solved in G and gives that $b = a^{-1}$ in G. Now let $a \in S$. The equation ab = e contains two elements, a and e, that are in S. If b were not in S then ab would not be in S, and since ab = e, an element of S, then b also must exist in S by closure.

Either I am reinventing the wheel, or this is a circular argument. Once again I suspect I am over-thinking this. One of my cardinal sins.

-Dan

5. Let s be a member of the subset, S. Since S is closed, $s^n$ is in S for all positive integers s. Since S is finite, we must have $s^i= s^j$ for some positive integers, i and j. We can, without loss of generality, assume i< j. Show that $s^{i- j}= e$.

6. let s be in S. if we consider the map g-->sg from S-->S (since we have closure on S, sg is again in S), it is not hard to show that it is bijective:

suppose sg = sh. then (since both sg and sh lie in G), we know that s^-1(sg) = s^-1(sh) in G, hence g = h. thus g-->sg is injective and since S is finite, it is bijective.

but if it is bijective (and thus surjective), that means we have some element in S, say s1, with s(s1) = s. again, in G we can multiply by s^-1 to get s1 = e.

this shows e is in S. now for inverses. since g-->sg is surjective, and e is in S, for some g in S we must have sg = e. but by the uniqueness of inverses in G, g must be s^-1.

thus S has the identity and inverses, so S is a subgroup.

7. Originally Posted by HallsofIvy
Let s be a member of the subset, S. Since S is closed, $s^n$ is in S for all positive integers s. Since S is finite, we must have $s^i= s^j$ for some positive integers, i and j. We can, without loss of generality, assume i< j. Show that $s^{i- j}= e$.
i like this idea, it shows in one fell swoop that <s> must be contained in S, for any s in S. i think you should have $s^{j-i}= e$ though, since i < j, and you have not shown that $s^{-n}$ is in S. once you know that $s^{j-i}=e$, then $s^{j-i-1}=s^{-1}$.

8. Thanks to you all. I got it.

-Dan

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