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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?