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We will show that a quasinormal operator is necessarily subnormal but not vice versa. Thus the normal operators is a proper subfamily of quasinormal operators, which in turn are contained by the subnormal operators. To argue the claim that a quasinormal operator is subnormal, recall the following property of quasinormal operators:
'''Fact:''' A bounded operator ''A'' is quasinormal if and only if in its [[polar decomposition]] ''A'' = ''UP'', the partial isometry ''U'' and positive operator ''P'' commute.<ref name="ConwayOlin1977">{{citation|author1=John B. Conway|author2=Robert F. Olin|title=A Functional Calculus for Subnormal Operators II|url=https://books.google.com/books?id=yQXUCQAAQBAJ|accessdate=15 June 2017|year=1977|publisher=American Mathematical Soc.|isbn=978-0-8218-2184-8|page=51}}</ref>
Given a quasinormal ''A'', the idea is to construct dilations for ''U'' and ''P'' in a sufficiently nice way so everything commutes. Suppose for the moment that ''U'' is an isometry. Let ''V'' be the unitary dilation of ''U'',
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