Is a bijective map between two groups always an isomorphism? Does a bijective map between two groups always produce an isomorphism?
I am trying to find a bijective map between two groups which does not preserve the group operations.
I have found a bijection $f(g)=g^3$, where $f:\mathbb{R}\rightarrow\mathbb{R}$, 
defined by $f(g+h)=(g+h)^3$. 
But $f(g + h) \neq     $
$ f(g)+f(h) \\=g^3+h^3$
 A: For a group $G$ with $\lvert G\rvert > 1$, we can find a bijection from $G
\to G$ that's  not a homomorphism.
Since the order of $G$ is at least $2$, there is a $g\in G$ with $g\neq 1_G$.
Now the map $f_g : G \to G$
$$
f_g : x \mapsto
\begin{cases}
g & \text{if } x = 1_G,\\
1_G & \text{if }x = g,\\
x & \text{otherwise}
\end{cases}
$$
is a bijection, but not a homomorphism (since $f_g(1_G) \neq 1_G$).
A: Interestingly, there are a few groups where any bijective map from the group to itself which preserves the identity element are automorphisms (ie, isomorphisms of the group with itself).
These groups are precisely $\{1\}$, $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/3\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$.
It is easy to check that these groups satisfy the above property, and that these are the only ones follows from my answer at Is a Bijection From a Group to Itself Automatically an Isomorphism If It Maps the Identity to Itself?
A: This is by means of azimut's answer. Black is group G. Orange is x. 

A: A maybe more "natural" example is the following map: if $G$ is not trivial, pick $g\in G,g\neq 1_G$.
Then the map $$\ell_g:x\in G\mapsto gx\in G$$ is bijective (with inverse $\ell_{g^{-1}}$), but not a group homomorphism, since $g=f(1_G)=f(1_G ^2)\neq f(1_G)^2=g^2$ (otherwise, by simplification by $g$, we would have $g=1_G$)
