Given $(G,*)$ a group. Show the function $f: G \to G$ defined as $f(a)=a^{-1}$ is an isomorphism if and only if G is abelian.
I am having troubles showing both ways of the proof. I know I have to show that $a*b=b*a$ for the first part, but do I have to suppose that the function is bijective or just use that $f(a*b)=f(a)*f(b)$?
For the second part, knowing that G is abelian, do I have to show that the function is bijective and $f(a*b)=f(a)*f(b)$ is true?
On
For any group $G$, the map
$f(g) = g^{-1}, \; g \in G, \tag 1$
is always a bijection: the quickest way to see this, I think, is to note that
$f^2(g) = f(f(g)) = f(g^{-1}) = (g^{-1})^{-1} = g, \tag 2$
so that
$f^2 = \mathbf 1_G, \tag 3$
the identity mapping on $G$:
$\mathbf 1_G(g) = g, \; \forall g \in G; \tag 4$
thus, from (2), (3),
$f(g) = f^{-1}(g), \; \forall g \in G, \tag 5$
that is, $f$ is its own inverse. Since any invertible map from any set to itself is bijective, so is $f:G \to G$.
If now $G$ is abelian,
$f(g_1 g_2) = (g_1 g_2)^{-1} = g_2^{-1} g_1^{-1} = g_1^{-1} g_2^{-1} = f(g_1) f(g_2), \tag 6$
which shows that $f$ is a homomorphism; this combined with the bijectivity of $f$ indicates it is an isomorphism.
Conversely, if $f$ is an isomorphism, then
$g_2^{-1} g_1^{-1} = (g_1 g_2)^{-1} = f(g_1 g_2) = f(g_1) f(g_2) = g_1^{-1} g_2^{-1}, \tag 7$
whence
$g_1 g_2 = (g_2^{-1} g_1^{-1})^{-1} = (g_1^{-1} g_2^{-1})^{-1} = g_2 g_1, \; \forall g_1, g_2 \in G, \tag 8$
which shows that $G$ is abelian indeed.
You can observe that for every $a,b\in G$
$(a*b)*(b^{-1}*a^{-1})=a*(b*b^{-1})a^{-1}=$
$=a*1*a^{-1}=a*a^{-1}=1$
so for the unicity of inverse you have that
$(ab)^{-1}=b^{-1}a^{-1}$
For every $a,b\in G$
$f(a*b)=(a*b)^{-1}=b^{-1}*a^{-1}=$
$=a^{-1}*b^{-1}=f(a)*f(b)$
so the map is a morphism.
$f(a)=1 \to a^{-1}=1$ and so $a=1$
so the map is injective
If $a\in G$ than
$f(a^{-1})=(a^{-1})^{-1}=a$
so the map is also surjective
Conversely, if $f$ is an isomorphism than
$a*b=f((a*b)^{-1})=$
$=f(b^{-1}*a^{-1})=f(b^{-1})*f(a^{-1})=b*a$