# Prove that $\phi$ is an isomorphism if and only if $G$ is abelian.

Let $G$ be a group and let $\phi:G \rightarrow G$ be defined by $\phi(g)=g^{-1}$. Prove that $\phi$ is an isomorphism if and only if $G$ is abelian.

My attempt.

(Forward direction)

Suppose $\phi$ is isomorphic. Then $\phi$ is:

$i)$ Injective. Implying that $\phi(g_1)=\phi(g_2), g_1=g_2$

$ii)$ Surjective. Implying that $a^{-1} \in G$ and $\phi(a)=a^{-1}$

$iii)$ $\phi(g_1g_2)$ = $\phi(g_1)\phi(g_2)$

from ($iii)$, $\phi(g_1g_2) =\phi(g_1)\phi(g_2)= g_1^{-1} g_2^{-1}$

=$\phi(g_2)\phi(g_1^{-1})\phi(g_1)\phi(g_2)\phi(g_2^{-1})\phi(g_1)= g_2^{-1}g_1g_1^{-1} g_2^{-1}g_2g_1^{-1}= g_2^{-1}g1^{-1} = \phi(g_2)\phi(g_1)$

Hence, it is abelian.

(backwards direction)

Suppose $\phi$ is abelian. Then $\phi(g_1 g_2) = g_2^{-1} g_1{^-1}$

To show injectivity: Let $\phi(g_1g_2) = \phi(g_2g_1)$

$g_1^{-1}g_2^{-1}= g_2^{-1}g_1^{-1}$

$g_1g_1^{-1}g_2^{-1}g_2=g_2g_2^{-1}g_1^{-1}g_1$

$e_1 = e_2$

Surjectivity follows by definition.

Finally, from similar to the forward direction, $\phi(g_1g_2)=\phi(g_1)\phi(g_2)$

• I do not think that the first direction is coherent. You should start with let $g_1,g_2 \in G$ and show that $g_1g_2=g_2g_1$, which you haven't done. Also, the "injective" statement doesn't make sense. It should be $\phi(g_1)=\phi(g_2) \implies g_1=g_2$. This is not the only way to do the proof, but as written, I can't follow the logic. – Andres Mejia Nov 2 '16 at 21:18

Assume $G$ is abelian and let $a,b \in G$. $\phi(ab) = (ab)^{-1} = b^{-1}a^{-1} = a^{-1}b^{-1} = \phi(a)\phi(b)$. $\phi$ is obviously surjective by construction. Suppose that $\phi(a) = \phi(b)$, then $a^{-1} = b^{-1} \longrightarrow aa^{-1}b = ab^{-1}b \longrightarrow b = a$. Thus $\phi$ is injective.

Now suppose that $\phi$ is an isomorphism. Then $\phi(ab) = \phi(a)\phi(b) = a^{-1}b^{-1}$ implies that $(ab)^{-1} = b^{-1}a^{-1} = a^{-1}b^{-1}$. Thus $G$ is abelian.

$\phi$ itself is one-to-one and onto without the condition that $G$ is abelian.

Suppose $\phi(g_1)=\phi(g_2)$ where $g_1,g_2\in G$
We have $g_1^{-1}=g_2^{-1}$
Since the inverse of an element is unique in a group,
$g_1=g_2$ which implies $\phi$ is one-to-one.

Let $y\in G$
Then $y^{-1}\in G$ where $\phi(y^{-1})=(y^{-1})^{-1}=y$
Thus $\phi$ is onto.

So the question now become Prove that $\phi$ is a homomorphism iff $G$ is abelian.

Suppose $\phi$ is a homomorphism.
Let $g_1,g_2\in G$
Then $\phi(g_1g_2)=\phi(g_1)\phi(g_2)$
So $(g_1g_2)^{-1}=g_1^{-1}g_2^{-1}=(g_2g_1)^{-1}$
Again since the inverse of an element is unique,
$g_1g_2=g_2g_1$, we conclude $G$ is abelian.

Suppose $G$ is abelian. Let $g_1,g_2\in G$
Then $\phi(g_1g_2)=(g_1g_2)^{-1}=g_2^{-1}g_1^{-1}=g_1^{-1}g_2^{-1}=\phi(g_1)\phi(g_2)$ We conclude that $\phi$ is a homomorphism.