Does there exist an onto group homomorphism from
- $\mathbb{R} \to \mathbb{C}$
- $\mathbb{C} \to \mathbb{R}$?
For $2$, consider the map $f(a+bi)=a$. Then for any $a\in \mathbb{R}$ we have $a+bi\in \mathbb{C}$ such that $f(a+bi)=a$.
I am stuck on the first one though. Please help.
Use the axiom of choice to obtain a Hamel basis $(x_i)_{i \in I} $ of $\Bbb {R} $ (i.e., a basis of $\Bbb {R} $ as a $\Bbb {Q} $ vector space). It is not too hard to see that this implies that the index set $I $ has the cardinality of the continuum.
Likewise, we can choose a Hamel basis $(y_j)_{j \in J} $ of $\Bbb{C} $ which again has the cardinality of the continuum (alternatively, as noted in the comments, a suitable choice of the $(y_j)_J $ is $(x_\ell)_{\ell \in I} \cup (i x_\ell)_{\ell \in I} $, which has the same cardinality as before since $I$ is infinite).
Choose a bijection $\sigma : I \to J $ and define
$$ \tau : \Bbb {R} \to \Bbb {C}, \sum_ i c_i x_i \mapsto \sum_i c_i y_{\sigma (i)}. $$
It is not hard to see that this yields a group isomorphism between $\Bbb {R} $ and $\Bbb {C} $.