I envisioned giving equal probability to each of the subgroups of cheese considered up to eat quality ... Let me try that again: I envisioned giving equal probability to each of the subgroups of G considered up to equality (as opposed to conjugacy). (I was speaking into my smart phone, and it changed my mathematical sentence into a curious culinary one.) Jim On Friday, May 4, 2018, Dan Asimov <dasimov@earthlink.net> wrote:

The problem was not what I wrote way below: Obviously with a finite number of groups of order n, one can easily pick one "at random".

But how about the subgroups? Are we picking one from *all* subgroups, including all those conjugate to one subgroup?

That method will tend to decrease the probability of a normal subgroup, since every *other* subgroup will get counted once for each of its conjugates but a normal subgroup has only one conjugate.

On the other hand, picking just one subgroup from each conjugacy class will increase the probability of a normal subgroup.

So: What is the probability measure being considered?

—Dan

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(That is, for large n, if you pick a group G of order n uniformly at random, and then pick a subgroup H of G uniformly at random, the probability that H is a normal subgroup of G goes to 0 as n goes to infinity.)

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Jim Propp wrote: ----- for large n, if you pick a group G of order n uniformly at random -----

I wonder how to define a probability on groups of order n.

And in particular: If there are several ways to do this, do they result in the same probabilities?

—Dan

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