- #1

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So I came up with two ways of writing the solution and am wondering whether they are equivalent and which one is preferable:

1) Let ##a,b \in G##. Then ##\phi (ab) = \phi(a)(b) \implies (ab)^{-1} = a^{-1}b^{-1} \implies b^{-1}a^{-1} = a^{-1}b^{-1} \implies ba = ab##.

2) Let ##a,b \in G##. Then ##ab = (b^{-1}a^{-1})^{-1} = \phi(b^{-1}a^{-1}) = \phi(b^{-1}) \phi(a^{-1}) = b a##