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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?