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Since is isomorphic to, there exists an isomorphism.
Because is isomorphic to, there exists an isomorphism.
Because Im is isomorphic to Ker, there exists an isomorphism.
Since ImQ is isomorphic to KerL, there exists an isomorphism (J operatorname{Im} Qrightarrowoperatorname{Ker} L).
(2) is isometrically isomorphic to.
Then, is isomorphic to by defining by.
Now (T_1 ) is isomorphic to (Q).
All cyclic groups are isomorphic to one of these.
Set (J_1 = xR) which is isomorphic to (D).
It is isomorphic to the Jacobson algebra [9].
Clearly, u k m 1 is isomorphic to ℝ4T.
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CEO of Professional Science Editing for Scientists @ prosciediting.com