Does every group homomorphism from $(0,\infty),\times)$ to $(\mathbb{R},+)$ send $1$ to $1$? I just have some true/false questions I am revising with and I'm not sure about this.
Let $f:((0,\infty),\times)\to(\mathbb{R},+)$ be a group homomorphism, then $f(1)=1\tag{1}$
I know that a group homomorphism is when you have 2 groups $(G,*)$ and $(H,\odot)$ such that a function $f:G \to H$ gives $f(a*b)=f(a) \odot f(b)$ for all $a,b \in G$
If someone could explain that (or give me hint), it'd be awesome!
Okay actually as I was typing this I may have had an epiphony!
I think that (1)is  false because $1\times 1 \neq 1+1$
I also have,
Let $f:((0,\infty),\times)\to(\mathbb{R},+)$ be a group homomorphism, then $f(\frac{1}{2})=-f(2)\tag{2}$
But please don't tell me the answer to this one! I'll try figure it out based on feedback from the first one and go from there
 A: Clearly $(1)$ is not true. In general, given $f:G\rightarrow H$ group homomorphism, we know that the identity element $e_G\in G$ maps to the identity element $e_H \in H$.
You just need to identify these elements in your groups and substitute in
$$f(e_G)=e_H. \tag{1}$$
Then it should be easy to figure out $(2)$, since
$$f(1)=f\left(\frac{1}{2}\cdot 2\right) = f\left(\frac{1}{2}\right) + f(2).$$
A: In general, if $(G,\star)$ and $(H,\perp)$ are two groups and $f \, : \, G \, \rightarrow \, H$ a group homomorphism, then $f$ satisfies :
$$ \forall (x,y) \in G^{2}, \; f(x \star y) = f(x) \perp f(y) \tag{$\clubsuit$} $$
Letting $x=y=e_{G}$ in $(\clubsuit)$, we have :
$$ f(e_{G}) = f(e_{G}) f(e_{G}) $$
Since $f(e_{G}) \in H$, it has an inverse $\big( f(e_{G}) \big)^{-1}$ in $H$. Multiplying by $\big( f(e_{G}) \big)^{-1}$, we obtain : 
$$f(e_{G}) = e_{H}. \tag{1}$$
As a consequence, let $x \in G$ and $y=x^{-1} \in G$. $(\clubsuit)$ gives :
$$ f(x \star x^{-1}) = f(e_{G}) = e_{H} = f(x) \perp f(x^{-1}) $$
Therefore :
$$ \forall x \in G, \, \big( f(x) \big)^{-1} = f(x^{-1}). \tag{2}$$
Application : let $G = (]0,+\infty[,\times)$ and $H = (\mathbb{R},+)$. Here, $e_{G} = 1$ and $e_{H} = 0$. $(1)$ writes :
$$ f(1) = 0. $$
And $(2)$ with $x=2 \in G$ writes :
$$ f(\underbrace{2^{-1}}_{\text{inverse in G}}) = f \Big( \frac{1}{2} \Big) = \underbrace{\big( f(2) \big)^{-1}}_{\text{inverse in H}} = - f(2). $$
A: Use the same idea to view $\frac 1 2$ and $2$. 
Try to view this 
$$f(\frac  1 2)=f(\frac 1 4.2)=2\frac 1 4\tag 1$$
but $$ -f(2)=-f(\frac 1 2.4)=4\frac 1 2\tag 2$$ which is not same.
