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abstract algebra

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Let M and N be two Normal subgroups of G. Prove that
(1) MN=NM
(2) MN is a normal subgroup of G
(3) if M ^ N = {e} then MN/N and M are isomorphic ( ^ means "and")

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Proof:
(1) For each element in MN, it has form mn, where m is in M and n is in N.
Since N is normal, then mnm^(-1) is in N. So mn = (mnm^(-1))m is in NM. Thus MN is a subset of NM.
For each element in NM, it has form nm, where n is in N and m is in M.
Since M is normal, then nmn^(-1) is in M. So nm = (nmn^(-1))n is in MN. Thus NM is a ...

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