# I Ways of writing a logical argument

#### Mr Davis 97

Let $G$ be a group. Suppose that the map from $G$ to itself defined by $\phi (g) = g^{-1}$ is a homomorphism. Prove that $G$ is abelian.

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 $ab = (b^{-1}a^{-1})^{-1} = \phi(b^{-1}a^{-1}) = \phi(b^{-1}) \phi(a^{-1}) = b a$

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#### fresh_42

Mentor
2018 Award
It doesn't matter, they are both valid. I found the first one easier to read (you've forgotten one $\phi$), but this is a matter of taste.

The basic idea is contained in both of them: inversion reverses the order whereas a homomorphism preserves the order. Both at the same time forces commutativity.

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