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I didn't understand how A non-empty subset H is a sub-group of a finite group G  if and only if for all a belongs to H, b belongs to H,  a*b belongs to H.

For example, (⎨1, 1/2 and 2⎬, *) is a group on multiplication but H = ⎨1, 1/2 ⎬ is not a sub-group. This is because there is no inverse for 1/2 in H.

However, the PDF linked to on the Gate CSE site under 'Set Theory and Algebra' says:

Could someone please tell me what this means/how this is true?

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The reasoning you provided is correct and the group that you have chosen will have only one sub group i.e {1} only.
Follow the standard definition of group: Closure , Associativity , identity ,inverse.
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according to LaGrange's theorem you cannot form a subgroup of order 2 with a group of order 3

there can be just one subgroup and that's identity element

Your set = { 1, 2, 1/2 }

on * operator 2 * 2 = 4 not in Set. Hence not closed => not Group => no question of finite subset.

Besides, Necessary condition for a subgroup is,

order(Subgroup) = order(Group) / k , k = {1,2,3,...}

Here order(G) = 3

Hence order(S) can be only 1 as 1 = 3 / 3

Example you chose is, Subgroup = { 1, 1/2 }, hence order(S) = 2

which does not divide 3. Hence not subgroup by LaGrange's theorem.

edited

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