Q:

Structure transport using a bijection

Let $(G,\cdot)$ be a group. Consider a set $H\neq \emptyset$ such that there exists a bijective function $f:G\to H$ with the property that $f(x\cdot y)=f(x)\cdot f(y)$,$\forall x\in G$. Is it true then that $(H,\cdot)$ is also a group?

I am not able neither to prove this nor disprove it, but I will provide how I came up with this.    
I started from the well-known fact that if $(G,\cdot)$ is a group and we have a set $M\neq \emptyset$ such that there exists a bijective function $f:G\to M$, then the following assertions are true:     
a) There exists a unique binary operation $*$ such that $f(x\cdot y)=f(x)*f(y)$,$\forall x,y\in G$, and this operation is $\forall a,b \in M, a*b=f(f^{-1}(a)\cdot f^{-1}(b))$.   
b)$(M,*)$ is a group isomorphic to $(G,\cdot)$.     
After seeing this result, I wanted to find out if there is some way to have the same binary operation on both sets instead of defining a new one. This is how I came up with the result from the beginning of my post. Intuitively, it seems true, because $f$ is something like a group isomorphism and I think that it should "transport" $(G,\cdot)$'s structure to $(H,\cdot)$. 
However, I can't prove this.     
Should the result I thought up be false, please show me under what conditions we may obtain what I want.

A:

If $f$ is bijective, then $\forall y \in H$, exists a unique $x \in G$ such that $f(x)=y$. By property of $f$, we have:
i) $\bar{e}=f(e)$ is a neutral element of $H$ (where $e$ is a neutral element of $G$). In fact, $\forall y \in H$:
$$
y \cdot \bar{e} = f(x)\cdot f(e) = f(x \cdot e) = f(e) = \bar{e}
$$
ii) If $y_1,y_2,y_3 \in H$, then:
$$
y_1 \cdot (y_2 \cdot y_3) = f(x_1) \cdot (f(x_2) \cdot f(x_3)) = f(x_1)\cdot f(x_2\cdot x_3) = f(x_1\cdot(x_2\cdot x_3)) = f((x_1 \cdot x_2)\cdot x_3) = (y_1\cdot y_2)\cdot y_3
$$
iii) For all $y \in H$, exists a unique $x \in G$ such that $f(x)=y$. Then, $y^{-1}=f(x^{-1})$, where $x^{-1}$ is a inverse of $x$. In fact:
$$
y \cdot y^{-1} = f(x)\cdot f(x^{-1}) = f(x\cdot x^{-1}) = f(e) = \bar{e}
$$
Therefore, $(H,\cdot)$ is a group.

