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Y by felixgotti
Let be a submonoid of , where runs over is the index set. Let be the intersection of all possible submonoids of . Notice now we just need to prove . Let be the identity of notice that and in turn therfor has the identity of the first characteristic of a submonoid. Now we just need to prove that it is a monoid with respect to the operation of . Let by intersection rules and and notice is a submonoid in itself and thus the operation for every thus by intersection rules again. Thus is a submonoid in itself!
Let be a subgroup of , where runs over is the index set. Let be the intersection of all possible subgroups of . Notice now we just need to prove . Let be the identity of notice that and in turn therfor has the identity of the first characteristic of a subgroup. Now we need to prove that it is a subgroup with respect to the operation of . Let by intersection rules and and notice is a subgroup in itself and thus the operation for every thus by intersection rules again. Finally, we need to show that every element has an inverse. Let and let represent the inverse of . Notice if then it follows that although notice that all are subgroups of and thus and since for any then by intersection thus is a subgroup of
Let be a subgroup of , where runs over is the index set. Let be the intersection of all possible subgroups of . Notice now we just need to prove . Let be the identity of notice that and in turn therfor has the identity of the first characteristic of a subgroup. Now we need to prove that it is a subgroup with respect to the operation of . Let by intersection rules and and notice is a subgroup in itself and thus the operation for every thus by intersection rules again. Finally, we need to show that every element has an inverse. Let and let represent the inverse of . Notice if then it follows that although notice that all are subgroups of and thus and since for any then by intersection thus is a subgroup of
This post has been edited 3 times. Last edited by Thayaden, Sep 16, 2024, 6:31 PM