>>13734354>>13734352>>13734062If you're wondering about the replies to this
>>13733754, most of it is really simple. Just grab a book on group theory (no prereq other than understanding how sets and proofs work) and you could pretty much get what's going on.
I'm gonna give a quick explanation, but some of it may go over your head a bit, and I'm omitting a lot, but that's okay. You can fill the gaps on your own easily.
Z = the integers = {...,-3,-2,-1,0,1,2,3,...}
You can take the the integers with the operation of addition +, and you have a group (look up the definition). You can take subsets of it that turn out to be also groups under addition, for example, 2Z (the even numbers), or 3Z (the multiples of 3), or 4Z (the multiples of 4), etc.
If you remember from primary school, you were taught that "even + even = even, odd + even = odd, odd + odd = even". Well, the set {even, odd} is a group under this new operation of addition, and you can get it from Z by "quotienting out" by the even numbers 2Z, and you get Z/2Z = {even, odd} = {2Z, 1+2Z}.
You could do the same with 3Z and get Z/3Z which has 3 elements, and in general Z/nZ has n elements. So for any n, you can get a quotient that is of size n, and so you can get quotients as big as you want. But there are no infinite quotients of Z (except itself).
Then the anon goes on to ask if there are infinite groups where we can have subgroups as big as we'd like that are normal (special property, those are the subgroups we can quotient by), but none that are infinite normal subgroups.
Then another anon gave an example of the rational numbers with a denominator that is a power of 2, but you take them "mod 1". Mod 1 means that whenever the sum goes over 1, you wrap back around to keep the result between 0 and 1. For example, 3/4 + 7/8 = 13/8, but then you wrap around by subtracting multiples of 1, and you get 5/8.
Its only (normal) subgroups are the numbers with a denominator of at most 2^k for a given k.
