Let $G$ be a topological group the underlying set of which is infinite (e.g., $(\mathbb{R}\,;+)$ or $(\mathbb{Z}\,;+)$), and let $H$ be a topological group the underlying set of which is finite (e.g., the group $P(n\,;\mathbb{R})$ of $(n\times n)$ permutation matrices).
My questions are:
- Is it possible to have a non-trivial group homomorphism $\phi:G\longrightarrow H$?
- If so, is it possible to have a $\phi$ such that the mapping $G\ni x\mapsto \phi(x)\in H$ is continuous?
Specifically, I am interested in the case $G=\mathbb{R}$ and $H=P(n\,;\mathbb{R})$.
Thank You.
Best Answer
Yes, you can have nontrivial homomorphisms. For instance, from $$(\mathbb Z, +)$$ to $$ (\mathbb Z/3\mathbb Z, +) $$ you can send $n$ to $n \bmod 3$.
Consider the matrix $$ A = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} $$ Since $A^3 = I$, the map that sends $n \in (\mathbb Z, +)$ to $A^n$ in $P(3; \mathbb R)$ is essentially the same map as in the first example, except that this time, the $\mathbb Z / 3 \mathbb Z$ is a subgroup of the matrix group.
As for maps from $\mathbb R$ to $P(n; \mathbb R)$: the homomorphic image of a connected topological group will still be connected, and the only connected subgroup of $P(n; \mathbb R)$ is the trivial one. (Indeed, this remark applies to any codomain that's discrete, as @Clement notes .)