Groovy Sort List

I am posting a simple example on how to sort a list in groovy because the examples google knows about aren't what I was looking for. With some deep digging I was able to find a clue that eventually solved my problem.

It's real easy to sort a list of numbers

assert [1,2,3,4] == [3,4,2,1].sort()

Or even strings

assert ['Chad','James','Travis'] == ['James','Travis','Chad'].sort()

But this was my example

class Person {
  String id
  String name
}

def list = [new Person(id: '1', name: 'James'),new Person(id: '2', name: 'Travis'), new Person(id: '3', name: 'Chad')]

list.sort() returns James, Travis, Chad

The solution is ridiculously simple (not that I thought the previous sort would work; I have to be realistic; groovy can't do everything for me).

list.sort{it.name} will produce an order of Chad, James, Travis.

In the previous example note the use of the sort closure sort {} verses the sort() method.

Now I am not sure, off the top of my head and without a Groovy book handy, the simplest way to sort case insensitive.

assert ['a1','A1'] == ['A1,'a1'].sort{fancy closure}