For groups, showing that a subset is closed under operation

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Mr Davis 97
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To show that a subset of a group is a subgroup, we show that there is the identity element, that the subset is closed under the induced binary operation, and that each element of the subset has an inverse in the subset.

My question is regarding showing closure. To show that the subset is closed under the operation, if we assume that ##a## and ##b## are elements of the subset , do we have to show that ##ab## is still in he subset or that ##ba## is also in the subset?
 
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Mr Davis 97 said:
To show that a subset of a group is a subgroup, we show that there is the identity element, that the subset is closed under the induced binary operation, and that each element of the subset has an inverse in the subset.

My question is regarding showing closure. To show that the subset is closed under the operation, if we assume that ##a## and ##b## are elements of the subset , do we have to show that ##ab## is still in he subset or that ##ba## is also in the subset?
If you show that ##ab## is in the subset for all possible combinations ##(a,b)##, does this include ##(b,a)\,##?
 
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