Genady

This was a challenging exercise. Give an example of an infinite field with a finite subfield. The simplest example I could come up with was a field of rational functions, \[\frac {a_n X^n+a_{n-1} X^{n-1}+...+a_0}{b_m X^m+b_{m-1} X^{m-1}+...+b_0}\] where all the coefficients a and b are from the field (0,1) as in the second post on the top of this thread,

They do, with \(b=0\) and a rational \(a\).

OK Let's take a different exercise. Show that any subfield of \(\mathbb C\) contains \(\mathbb Q\). It has to contain 0 and 1. From those, with addition and subtraction, it has to contain all \(\mathbb Z\). From there, with multiplication and division, it has to contain all \(\mathbb Q\). QED

Right. This set is not closed under usual addition. That is why it is not a field with respect to the usual addition.

Thank you. I don't think it fits in this exercise because 1 + 1 = 0 mod 2 while the exercise asks about "usual addition and multiplication of numbers".

I'd like to make textbook exercises, some as a refresher and others new to me, and I hope that mathematicians here will take a look and will point out when I miss something. Here is a first bunch. Which of the following subsets of \(\mathbb C\) are fields with respect to the usual addition and multiplication of numbers: (a) \(\mathbb Z\)? Not a field. E.g., no inverse of 2. (b) \(\{0,1\}\)? Not a field. Not even a group. (c) \(\{0\}\)? Not a field. No multiplicative identity. (d) \(\{a+b\sqrt 2; a, b \in \mathbb Q\}\)? Yes. (e) \(\{a+b\sqrt[3] 2; a, b \in \mathbb Q\}\)? Not a field. Can't get inverse of \(\sqrt[3] 2\). (f) \(\{a+b\sqrt[4] 2; a, b \in \mathbb Q\}\)? Not a field. Can't get inverse of \(\sqrt[4] 2\). (g) \(\{a+b\sqrt 2; a, b \in \mathbb Z\}\)? Not a field. E.g., no inverse of 2. (h) \(\{z \in \mathbb C : |z| \leq 1\}\) Not a field. No inverses when \(|z| \lt 1\).

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Anyway, it starts now here.

May I ask, in what senses they are not?

From Bug (engineering) - Wikipedia:

According to sources including wikipedia, it was a joke because the term had been in use already for a long time.

It was quite exciting and pleasant to watch our systems successfully crossing the date line in the time zone sequence first in Tokyo, then Singapore, London, and finally, NY. Being provided dinner and lodging in the downtown Manhattan, just in case.

You have shown that Since the answer to the question is Yes, you have shown that the alternating group has no subgroups of order half that of the group.

I think so.

Nice. So, you've proved that a simple group cannot have subgroups of order half that of the group. 😉 Because I know the answer.

... if it's a subgroup of a finite group G and contains exactly half the elements of G?

I'd rather file a complaint.

Perhaps. But I don't see this number in the text.

What do they mean by this?

It is also sufficient, right?

"It can be also generated by its other element" is a given.

G is a cyclic group of order n generated by an element g, G = {e, g, g2, ..., gn-1). It can be also generated by its other element h = gk, G = {e, h, h2, ..., hn-1}. How are the k and n related?

I think I'm missing something. If g1 = h1 g h1-1 and g2 = h2 g h2-1 then g1 ~ g and g2 ~ g, but it is not necessary that g1 ~ g2, which they should if this is an equivalence class. ?

As Baku native, I could be offended (I am not.)

Does it mean, g1 ~ g when for some h1, g1 = h1 g h1-1 ?