Why are all bijective group homomorphisms $\phi: G \to H$ automatically isomorphisms? That is, their inverses are also homomorphisms.
I have seen this fact mentioned before, but never a proof.
We need to show that $\phi^{-1}(ab) = \phi^{-1}(a) \phi^{-1}(b)$ for all $a, b \in G$… no idea where to start.
On
If $\phi:G \rightarrow H$ is a bijective group homomorphism, then we can show its inverse $\phi^{-1}: H \rightarrow G$ is also a group homomorphism.
Let $h,h' \in H$ be arbitrary. Then $\phi(g)=h$ and $\phi(g')=h'$ for some $g,g' \in G$. Since $\phi$ is a homomorphism, $\phi(g+g') =\phi(g) + \phi(g')=h+h'$.
Thus, $g+g'=\phi^{-1}(h+h')$.
Therefore, $\phi^{-1}(h) +\phi^{-1}(h')=\phi^{-1}(h+h')$.
On
Let $\varphi:G\to H$ be an bijective homomorphism. Then $\varphi^{-1}$ is a bijection. Let $h,h'\in H$. It remains to show that
$$\varphi^{-1}(hh')=\varphi^{-1}(h)\varphi^{-1}(h').$$
Since $\varphi^{-1}$ is a bijection, there exist $g,g'\in G$ such that
$$g=\varphi^{-1}(h), g'=\varphi^{-1}(h').$$
Now
$$\begin{align} \varphi^{-1}(hh')&=\varphi^{-1}(\varphi(g)\varphi(g'))\\ &=\varphi^{-1}(\varphi(gg'))\tag{1}\\ &=(\varphi^{-1}\circ\varphi)(gg')\\ &=\operatorname{id}_G(gg')\\ &=gg'\\ &=\varphi^{-1}(h)\varphi^{-1}(h'), \end{align}$$
where $(1)$ holds since $\varphi$ is a homomorphism.
On
The fact that $G$ and $H$ are groups is irrelevant. The result holds for all sets with a binary operation.
Let $A$ and $X$ be sets, with operations that we denote by multiplication. No further property is assumed.
A homomorphism $\varphi\colon A\to X$ is a map such that, for every $a,b\in A$, $\varphi(ab)=\varphi(a)\varphi(b)$.
If $\varphi$ is an isomorphism, that is, a bijective homomorphism, then also $\varphi^{-1}$ is a (bijective) homomorphism, hence an isomorphism.
Proof. Let $x,y\in X$. Then we have $$ \varphi(\varphi^{-1}(x)\varphi^{-1}(y))=\varphi(\varphi^{-1}(x))\varphi(\varphi^{-1}(y))=xy $$ and therefore $$ \varphi^{-1}(xy)=\varphi^{-1}(\varphi(\varphi^{-1}(x)\varphi^{-1}(y)))=\varphi^{-1}(x)\varphi^{-1}(y) $$
It's a useful exercise to prove that if $A$ is a group under the given operation and $\varphi\colon A\to X$ is an isomorphism, then also $X$ is a group.
The result extends to $n$-ary operations in the same way. You don't need to prove it for rings, fields, and even vector spaces, if you see them as sets with a binary operation + and a set of unary operations (the multiplications by a scalar).
On
Here is a one-line proof (so to speak).
$$\begin{align} \phi^{-1} (a b) &= \phi^{-1} \Big (\phi \big (\phi^{-1}(a) \big) \ \phi \big(\phi^{-1}(b) \big) \Big) \\ &= \phi^{-1} \Big (\phi \big (\phi^{-1}(a) \ \phi^{-1}(b) \big) \Big) \\ &= \phi^{-1}(a) \ \phi^{-1}(b) \end{align}$$
On
Let $$1_{G}\ \in\ G$$ $$1_{H}\ \in\ H$$ be the two groups’ respective identity elements and let $$\mu_{G}\ \colon\ G\times G\to G$$ $$\mu_{H}\ \colon\ H\times H\to H$$ be their respective multiplication maps (as functions of sets).
As $\phi$ is a homomorphism, $$\phi\left(1_{G}\right)\ = \ 1_{H}$$ $$\phi\ \circ\ \mu_{G}\ =\ \mu_{H}\ \circ\ \left(\phi\times\phi\right).$$ It follows (by bare definition) that \begin{align*} \phi^{-1}\left(1_{H}\right)\ &=\ \phi^{-1}\left(\phi\left(1_{G}\right)\right)\\ &=\ 1_{G}\\ \end{align*} and \begin{align*} \phi^{-1}\ \circ\ \mu_{H}\ &=\ \phi^{-1}\ \circ\ \left(\mu_{H}\ \circ\ \left(\phi\times\phi\right)\right)\ \circ\ \left(\phi^{-1}\times\phi^{-1}\right)\\ &=\ \phi^{-1}\ \circ\ \left(\phi\ \circ\ \mu_{G}\right)\ \circ\ \left(\phi^{-1}\times\phi^{-1}\right)\\ &=\ \mu_{G}\ \circ\ \left(\phi^{-1}\times\phi^{-1}\right) \end{align*} and thus that $\phi^{-1}$ is itself (by bare definition) a homomorphism, as claimed.
(The above is a shadow of the more general fact that a natural transformation is invertible iff it’s componentwise invertible.)
The inverse of a bijective map is bijective. We must show that for all $x$ and $y\in H$, $$\phi ^{-1}(x) \phi ^{-1} (y) = \phi ^{-1}(xy)$$Set $a = \phi ^{-1}(x), b = \phi ^{-1} (y)$ and $ c = \phi ^{-1} (xy)$. Since $\phi $ is a homomorphism, $$\phi (ab) = \phi (a)\phi (b) = x y = \phi (c) \Rightarrow \phi ^{-1}(x)\phi ^{-1}(y)=\phi ^{-1}(xy)$$