I am struggling to know exactly how to go about working out and proving that a mapping function is a group homomorphism for a particular example.
I have found that in order for the mapping to be a homomorphism, we must satisfy the condition
$f(x*y)=f(x)*f(y)$, where $f:G\to H$ is the function and this must be true for all $x,y\in G$.
With this in mind, we shall look at a simple example and ask,
is this a group homomorphism: $f:\mathbb Z \to \mathbb Z, x\longmapsto -2x$?
Here, it is obvious that the group law is addition and so it can be easily proven:
$f(x+y)=-2(x+y)=-2x-2y=f(x)+f(y)$ and so it is a group homomorphism.
For the example which I am having trouble with, the mapping is defined as
$g:\mathbb Z\to \mathbb R^*, x\longmapsto 2^{x^2}$.
The bit I am struggling with is knowing which group law to apply my test on since $\mathbb Z$ has the group law of addition whereas $\mathbb R^*$ has the group law of multiplication, so should it be addition or multiplication? Thanks!
This is definitely one of the hurdles in group theory! You can clarify the map by accompanying each set with the group operation: thus $g$ is the map $$g:\left\{\begin{array}{ccc} (\Bbb Z,+) & \rightarrow & (\Bbb R^*,\cdot) \\ x & \mapsto & 2^{x^2} \end{array}\right.$$ Therefore you want to check if $$\forall x,y\in\Bbb Z,\quad g(x+y)=g(x)\cdot g(y).$$ Notice how the symbol for the operation changes when going from inside the parentheses to outside.
Another remark: $(\Bbb R^*,+)$ is not a group, as it does not have a neutral element :)