# Non-trivial group homomorphism from an infinite group to a finite group

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:

1. Is it possible to have a non-trivial group homomorphism $\phi:G\longrightarrow H$?
2. 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.

• What does the $\Bbb R$ mean in $P(n;\Bbb R)$? – Arthur Apr 21 '15 at 11:34
• It means that the matrix elements are real. – SepulzioNori Apr 21 '15 at 11:42
• Shouldn't the elements be just $0$ or $1$ anyways? – Arthur Apr 21 '15 at 12:29
• I think so, but, since I am not sure that this is the most general definition of a permutation matrix (I am approaching this subject for the first time), I preferred to specify it. – SepulzioNori Apr 21 '15 at 12:40

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 .)

• Ok, and what about $\phi:\mathbb{R}\longrightarrow P(n\,;\mathbb{R})$? – SepulzioNori Apr 21 '15 at 11:46
• @Ilcapitano, $\phi(\mathbb{R})$ must be connected (if $\phi$ is continuous) and $P(n,\mathbb{R})$ is discrete (finite subspace of a Hausdorff space) so in this case there is a unique choice for $\phi(\mathbb{R})$. – Clément Guérin Apr 21 '15 at 11:47
• @ClémentGuérin: You are a absolutely right about a continuous $\phi$ (actually I should have realized it immediately), nevertheless, the case of a generic group homomorphism seems to me to be still open. – SepulzioNori Apr 21 '15 at 12:38

There is no non-trivial homomorphism from $\mathbb{R}$ to a finite group. The key fact is that $\mathbb{R}$ is divisible, meaning that every natural number $n$ and any $x\in \mathbb{R}$, there is a $y\in \mathbb{R}$ with $ny = x$.

Divisibility clearly implies a non-trivial group is infinite. (See this MSE questions for a proof).

Now, if there is a non-trivial homomorphism $\mathbb{R}\rightarrow G$, then the image of $\mathbb{R}$ is a non-trivial divisible subgroup of $G$. By the previous paragraph, this means $G$ has an infinite subgroup, so $G$ is infinite.