Proving a nonempty finite subset of a group is a subgroup

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  • #1
eq1
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Homework Statement



Let a non empty finite subset H of a group G be closed under the binary operation * on G. Show that H is a subgroup of H.

2. Relevant Definitions

Group Properties:
G1: a*(b*c)=(a*b)*c for all a,b,c in G
G2: e*x=x*e=x for all x in G
G2: if x is in G then x' is in G and x*x'=x'*x=e

A subset H of group G is a subgroup of G iff:
S1: the identity element of G, e, is in H
S2: the binary operation on G is closed on H
S3: if x is in H, then x' is in H and x*x'=x'*x=e


The Attempt at a Solution



So I know

H != {}
H is finite
H is closed, so I get the second property of a subgroup for free.
H is subset of the elements of G and inherits the operation.

If H only has one element, I only need to show that that element is the identity of G. I think I can do this pretty easy with the closure property.

If H has more then one element here is my strategy,

Assume x is in H and for every y in H x*y != e.
This breaks associativity on G which is a contraction since G is a group.
Therefore given x in H for some y in H x*y=e which means y is x's inverse and since it
is in H and since H is closed e is there too.

The reason I think this is the way to go is because I drew tables for a couple of binary operations for a small set that was a group. Then I added some elements around it with a new pseudo-identity element over the larger wanna-be group. Associativity broke on the larger operation, which means it couldn't have been a group.

So it seems like I should be able to sketch a proof around this concept. Given H is closed and associative and assume it does not contain the identity of G. This breaks associativity of G which is a contradiction. But I am not really getting anywhere.

Is this approach even valid? I am starting to suspect not. I am pretty stuck. Could someone perhaps suggest another direction? I am not really taking advantage of the finite piece of information. Maybe there is something there?
 

Answers and Replies

  • #2
morphism
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Suppose x is in H. Since H is closed under *, then x^n = x*x*...*x is in H for all n > 0.

But H is finite, so ...
 
  • #3
eq1
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Right, of course.

I think I have an outline of a proof that works.

H={x^n,for every n in Z}
Proof:
Assume there is a y in H where y != x^i. I can show this is a contradiction
by knowing x*c=y has a solution in G, and c cannot be a power of x, or y is a power
of x. But x^t*y is in H due to closure and for some t, x^t*y=e otherwise H is not
finite. But then x^t*y = x^t*x*c = e.
A contradiction, so c is a power of x, and y = x^i.

So now I know H is finite and has a generator x.
So x^0=e
And if |H|=n then x^i*x^(n-i)=e so there is an inverse for every element

H is a subgroup of G

I probably made that way harder than is had to be.
 
Last edited:
  • #4
morphism
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Why should it follow that x^t*y=e?

H certainly doesn't have to be cyclic. The purpose of my hint above was to let you invoke the finiteness of H to see that not all the x^n's are distinct, i.e. there must be an s and a t with, say, s>t, such that x^s = x^t. From this, one can deduce that e is in H and that x is invertible. But since x was arbitrary, this means every element has an inverse in H, so it's a subgroup of G.
 

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