Proving that $G = {2^i|i \in {\mathbb Z}}$ is a homomorphism from $(G,\cdot)\ to\ (Z,$ +). (Explanation of an existing solution)

Question

So here's the problem from my book:

Prove that $ f: G \rightarrow \mathbb Z\ defined\ through\ f(2^i) =i\ for\ all\ 2^i\ \in\ G$ is a homomorphism from $(G,\cdot)\ to\ (Z,$ +).

The proof in the book is the following: $f(2^i \cdot 2^j) =f( 2^{i+j})= i+j =f(2^i)+f(2^j).$

My questions:

1. Why is $f( 2^{i+j})= i+j$?

2. My understanding is that $2^i+2^j = i+j$ is impossible if $\ i \ne j$, so how is $=f( 2^{i+j})= i+j =f(2^i)+f(2^j)$ possible?

Detailed explanations are much appreciated, thank you everyone!

Answer 1

It's important to keep track of which binary operation is being performed in each group. In particular you've come up with $2^i + 2^j$ by applying the additive operation of $\mathbb{Z}$ to elements of the group $G$ but the binary operation on $G$ is multiplication, not addition so this is not allowed. Instead it should read $2^i \cdot 2^j$ which is what the textbook has done in the proof.

Answer 2

What is $G$? From the context, I guess it must be $\{2^k \mid k\in\Bbb Z\}$ with multiplication.

Then, we define one particular function, $f$, which assigns the exponent $k$ to the element $2^k$ of $G$
[e.g. $f(32)=5$ and $f(\frac18)=-3$].

Now, I hope it clarifies 1. [Since, by definition, $f(2^{whatever})=whatever$.]

For 2., indeed $2^i+2^j=i+j$ is impossible (even if $i=j$), but this fact has nothing to do with this problem, as we never equate any $f(k)$ to $k$.

The point is, that $f$ takes the multiplication of $G$ to the addition of $\Bbb Z$.

Note that the same definition works as a homomorphism $(\Bbb R_{>0},\cdot)\,\to\,(\Bbb R,+)$.