# How can I express a resulting discrepancy? (Proving $\varphi(x \in G) = x^{-1}$ is an automorphism iff G is abelian)

If anyone, after reading my question, can think of a better way to phrase the title, please feel free to edit it.

Let $G$ be an abelian group. Show that the mapping $\varphi: G \rightarrow G$ given by $\varphi(x) = x^{-1}$ is an automorphism of $G$. Show that if $G$ were not abelian, then $\varphi$ would not be an automorphism.

I'll start with the proof that $\varphi$ is not an automorphism if $G$ is not abelian:

Assume $G$ is not abelian. Then $\exists ~ g_1, g_2 \in G \ni g_1g_2 \neq g_2g_1$. We have:

$$\varphi(g_1g_2) =\varphi(g_1) ~ \varphi(g_2) = g_2\varphi(g_1 g_2) g_1 \varphi(g_2g_1) = \varphi(g_2g_1)$$

Since $g_1g_2 \neq g_2g_1$, but both map to the same element under $\varphi$, $\varphi$ is not one-to-one and thus not an automorphism.

Here's where I'm getting mixed up. In proving that $\varphi$ is monomorphic for abelian groups, I come up with this:

To show $\varphi$ is a monomorphism, we must show that $\varphi(x) = \varphi(y) \implies x = y$. Assume, then, $\varphi(x) = \varphi(y)$. This implies:

\begin{align}x^{-1} = y^{-1} &\implies x^{-1}x = y^{-1}x\\ &\implies yx^{-1}x = yy^{-1}x\\ &\implies y = x \end{align}

... but this didn't use the fact that $G$ is abelian at all, and seems as if it would hold even if $G$ were not abelian. There's clearly something wrong here, because above I proved the opposite of this. These are the two areas I think something might be going wrong:

1. I have to prove more to fully prove the implication here. If we let $P = [\varphi(x) = \varphi(y)]$ and $Q = [x = y]$, then, I've only shown that when $P$ is true, $Q$ is true, but not that whenever $Q$ is true, $P$ must be true (as per the truth table of implication). This will necessitate the use of the fact that $G$ is abelian. (I don't think this is it, since equality is reflexive.)

2. There's something amiss with $x^{-1}$ here, because the only guarantee that the inverse in $x^{-1}x$ is the same as the inverse in $xx^{-1}$ is the fact that $G$ is abelian. If this is the case, though, then what does $\varphi$ even mean? It seems to me like $\varphi(x) = x^{-1}$ isn't a sensible statement when something's left-inverse can be different from its right-inverse.

The issue is not injectivity. Instead, the problem is that $\phi$ is not a homomorphism if $G$ is not abelian.
Indeed, if $\phi(g_1g_2)=\phi(g_1)\phi(g_2)$ for all $g_1,g_2\in G$ then $$g_2^{-1}g_1^{-1}=g_1^{-1}g_2^{-1}$$ for all $g_1,g_2\in G$, which implies that $G$ is abelian.
• Oh, no! You know, while I was writing this question I thought exactly that. Then that [asshole] voice in the back of my mind said "Oh, no. The book said $\varphi$ was a homomorphism." I totally overlooked the problem and made a mess of things. Thanks very much. – AmagicalFishy Jan 25 '18 at 3:10