@MikaeldelaSalle: The order can be as large as $\exp((n\log(n))^{1/2})$. Take a disjoint union of prime cycles $(p_1)...(p_k)$ such that $p_1+\cdots+p_k \sim n$. Then this works for the first primes up to roughly $(n\log(n))^{1/2}$ and then $p_1\cdot \cdots \cdot p_k$ (the order of the permutation) is roughly $\exp((n\log(n))^{1/2})$. This was first observed by Landau.

@DenisT.: I think that this result on the maximal element order is already due to Landau. I would be interested to see the details of your approach!

The subgroup ${\mathbb Z}_p$ is open in ${\mathbb Q}_p$.

If $h$ is not continuous, then I do not see how one can use the separability of $G$ to get $H$ separable. Moreover, even if $H$ is separable, then $(1 \times h)(G)$ is not closed in $G \times H$. So the induced topology from the product is not polish (as it is not completely metrizable).

You are right. What I know is that it has a unique polish topology and every homomorphism to a polish SIN group is continuous.

I think that this is also the original proof of the statement I made (due to Kallman).

$SO(3)$ has also a unique SIN topology (not only totally bounded).

It is Example 1.5 in [C. Rosendal. Automatic Continuity of Group Homomorphisms; Bulletin of Symbolic Logic 15, no.2 (2009), 184-214.] with references to the work of Kallman and Thomas.

There is a discontinuous homomorphism from $SO(3,\mathbb R)$ to ${\rm Sym}(\mathbb N)$. This will likely give a non-complete topology.

I know a simpler paradox: At the $n$-th step add the balls from $n^2$ to $(n+1)^2-1$ and then remove the $n$-th ball. How many balls are left after infinitely many steps? This has nothing to do with probability.

You are right, I deleted this as it was nonsense.