It's true because if ab is unit then (ab)(b−1a−1)=(aa−1)(bb−1)=1⇒{aa−1=1bb−1=1⇒{a is unitb is unit\left(ab\right)\left(b^{-1}a^{-1}\right) = \left(aa^{-1}\right)\left(bb^{-1}\right) = 1 \Rightarrow \left\{ \begin{array}{l} aa^{-1} = 1 \\ bb^{-1} = 1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} a \text{ is unit} \\ b \text{ is unit} \end{array} \right.(ab)(b−1a−1)=(aa−1)(bb−1)=1⇒{aa−1=1bb−1=1⇒{a is unitb is unit
Answer: True.
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