# Showing that a set is a group, and a mapping is a group isomorphism

Let $(G,\cdot)$ be a group, $g \in G$.

For $a,b \in G$ define $a * b := a \cdot g^{-1} b$. Show that $(G,*)$ is a group with the neutral element $g$ and $f : (G,*) \rightarrow (G,\cdot), a \mapsto a \cdot g^{-1}$ is a group isomorphism.

In order to show that $(G,*)$ is a group, I need to show that $*$ is closed and associative, g is the neutral element and an inverse element exists for each $x \in G$.

(* is closed, associativity)

$*$ is closed because $\cdot$ is, $\forall x,y \in G: x * y = x \cdot g^{-1} \cdot y$, which is closed because $(G,\cdot)$ is a group. The same applies for associativity: $(x * y) * z = (x \cdot g^{-1} \cdot y) \cdot g^{-1} \cdot z = x \cdot (g^{-1} \cdot y) \cdot g^{-1} \cdot z) = x * (y * z)$.

(neutral element)

$x * g = x \cdot g^{-1} g = x \cdot e = x, g * x = g \cdot g^{-1} x = e \cdot x = x$

Is this right?

How do I show that there is an inverse element, so that $\forall x \in G \; \exists x': x * x' = g$?

In order to show that f is a group isomorphism, I need to show that it is a group homomorphism, and f is a bijection, right?

I already figured out how to show that f is surjective, but how do I show that f is injective?

To get the inverse of an element $a:$

$$a*b=ag^{-1}b=g\implies g^{-1}b=a^{-1}g\implies b=ga^{-1}g,$$

that is, the inverse of $a$ with respect to $*$ is $ga^{-1}g.$

To show that $f$ is injective:

$$f(a)=f(b)\implies ag^{-1}=bg^{-1}\implies a=b$$ (just multiply by the right by $g$ in the last step).

It is best to start from the end, i.e. first to show that $f$ has the isomorphism property (even if we don't know yet that $(G,*)$ is a group), that is $$\tag1 f(a*b)=f(a)\cdot f(b)\qquad\text{for all a,b\in G}$$ and $$\tag2 f\text{ is bijective}.$$ Once you have shown $(1)$ and $(2)$, the group properties are straightforward: $$f((a*b)*c)=f(a*b) f(c)=(f(a) f(b)) f(c)=f(a)(f(b) f(c))=f(a) f(b*c)=f(a*(b*c))$$ and hence by injectivity of $f$, $(a*b)*c=a*(b*c)$.

Or $a*f^{-1}(e)=f^{-1}(e)*a=a$ for all $a\in G$ because after application of $f$, we have $f(a)\cdot e=e\cdot f(a)=f(a)$. (And you can readily compute $f^{-1}(e)$ explicitly).

Moreover, the inverse is what it should be accoring to $f$, i.e. $a*a'=a'*a=f^{-1}(e)$ is equivalent to $f(a)f(a')=f(a*)f(a)=e$, i.e. we must have $a'=f^{-1}(f(a)^{-1})$.