2. If *f*{*ab*^{−1}) = *f*(*a*)*f*(*b*)^{−1} for all *a*, *b* ∈ *G*, then *f*(*e*) = *f*(*aa*^{−1}) = *f*(*a*)*f*(*a*)^{−1} = *e*', *f*(*b*^{−1}) = *f*(*eb*^{−1}) = *f*(*e*)*f*(*b*)^{−1} = *e*'*f*(*b*)^{−1} = *f*(*b*)^{−1} and *f*(*ab*) = *f*(*a*(*b*^{−1})^{−1}) = *f*(*a*)(*f*(*b*)^{−1})^{−1} = *f*(*a*)*f*(*b*) for all *a* and *b* ∈ *G*. The converse is clear.

4. *G* is abelian ⇔ *ab* = *ba* for all *a* and *b* ∈ *G*

⇔(*ab*)^{−1} = *b*^{−l}*a*^{−1} = *a*^{−1}*b*^{−1} for all *a*, *b* ∈ *G*

⇔*f*(*ab*)= *f*(*a*)*f*(*b*)

6. For any homomorphism *f* : ℤ → ℤ_{2}, *f*{*n*) = *nf*{1) for all *n* ∈ ℤ and hence

and *f*(1) = 1 ⇒ *f*(*n*)= 0 for all even *n* and

(since 1 1 1 = 0 in ℤ_{2})

8. Let *f* : ℤ → ℤ be a nontrivial endomorphism. Then, *f*{*n*) = *nf*{1) for all *n* ∈ ℤ and hence *f* ≠ 0 so that

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