0
$\begingroup$

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.

I've thought about this for a bit, and can't come to a satisfying conclusion.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – AmagicalFishy Jan 25 '18 at 3:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.