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2 adding quotient'

There are also many infinite order elements in the Haar measure sense:

Recall that a profinite group is compact, hence it has a probability Haar measure. If G is abelian and it has an element of order > n, then G has a either ℤ/pℤ for p > n+1 or ℤ/pkℤ for small p and suitably large k as a quotient. Thus under the assumption that G has elements of unbounded order we get that either the former is a quotient for infinitely many primes, or the latter is a quotient for a fixed prime and infinitely many powers.

Now the ratio of elements in each of the finite quotients of order bigger than $n$ tends to $1$. By standard arguments of measure theory, this implies that the probability measure of elements in $G$ of order bigger than $n$ is $1$ (when p tends to infinity or p is fixed and k tends to infinity). Taking intersection over all $n$'s give that the probability that an element will have an order bigger than any positive integer, i.e., infinite order, is $1$.

1

There are also many infinite order elements in the Haar measure sense:

Recall that a profinite group is compact, hence it has a probability Haar measure. If G is abelian and it has an element of order > n, then G has a either ℤ/pℤ for p > n+1 or ℤ/`pkℤ for small p and suitably large k. Thus under the assumption that G has elements of unbounded order we get that either the former is a quotient for infinitely many primes, or the latter is a quotient for a fixed prime and infinitely many powers.

Now the ratio of elements in each of the finite quotients of order bigger than $n$ tends to $1$. By standard arguments of measure theory, this implies that the probability measure of elements in $G$ of order bigger than $n$ is $1$ (when p tends to infinity or p is fixed and k tends to infinity). Taking intersection over all $n$'s give that the probability that an element will have an order bigger than any positive integer, i.e., infinite order, is $1$.