*author: niplav, created: 2024-01-26, modified: 2024-01-31, language: english, status: in progress, importance: 2, confidence: certain*

I've decided to learn some real math, not just computer scientist math.

Natural explations supersede proofs.

*—Evan Chen, “An Infinitely Large Napkin” p. 6, 2023*

Why do we need the fact that

`$p$`

is a prime?

If `$p$`

weren't a prime, then the operation is not closed (because there
are elements of the group that divide the group size), and therefore there
are elements that are not invertible. Take `$(ℤ/4ℤ)^{\times}$`

. Then
the element `$2$`

is not invertible:
`$2 \cdot 1=2, 2 \cdot 2=4 \text{ mod } 4=0, 2 \cdot 3=6 \text{ mod } 4=2$`

.

So the group operation is not closed, and also `$2$`

doesn't have
an inverse.

What are the identity and the inverses of the product group?

- Identity: The tuple that contains the identities of each group,
`$(1_G, 1_H)$`

- Inverses: The tuple that contains the element-wise inverses for
`$(g_1, h_1)^{-1}=(g_1^{-1}, h_1^{-1})$`

- (a) Rational numbers with odd denominators (in simplest form), where the operation is addition. (This includes integers, written as
`$n/1$`

, and`$0 = 0/1)$`

.

This is indeed a group.

- The identity element is 0.
- The operation
`$+$`

is associative. - Every element has an inverse, it's simply the negation of the element.

Now, is the `$+$`

operation closed?

$$\frac{a}{2n+1} + \frac{b}{2m+1}=\frac{(2m+1)a+(2n+1)b}{4mn+2n+2m+1}$$

It looks like the denominator must stay odd, but I'm not *sure* that's
necessary.

Assume `$(2m+1)a+(2n+1)b=u \cdot k$`

and `$4mn+2n+2m+1=l \cdot k$`

. Then
`$k$`

must be greater than or equal to three.

- (b) The set of rational numbers with denominator at most 2, where the operation is addition.

I assume we're excluding denominator zero.

Then we have identity (0), associativity and the inverse (again the negative). The operation looks pretty closed to me as well.

- (c) The set of rational numbers with denominator at most 2, where the operation is multiplication.

This set is not a group, because with the identity element `$1$`

the
number `$3$`

doesn't have an inverse.

- (d) The set of nonnegative integers, where the operation is addition.

This set is also not a group because it doesn't have the inverse for,
e.g., the number `$1$`

.

`$x \mapsto gx$`

is an injection: Assume there is a`$y$`

so that no`$x$`

so that`$gx=y$`

. Then let`$g^{-1}y=x'$`

(and ignore the suggestive naming). But then`$gg^{-1}y=gx'$`

and therefore`$y=gx'$`

. So such a`$y$`

can't exist.`$x \mapsto gx$`

is an surjection: Assume`$x \not =x'$`

. Assume also`$gx=y=gx'$`

. Then`$gx=gx'$`

. But then`$g^{-1}g=g^{-1}gx'$`

, so`$x=x'$`

.

A thing that tripped me up was that I then tried to prove that right
multiplication *isn't* a bijection—only to give up in confusion and
later find out that it is *also* a bijection. So much for suggestive
questions.

Let `$g$`

be the primite root modulo `$p$`

. Then the isomorphism between
`$ℤ/(p-1)ℤ \cong (ℤ/pℤ)^{\times}$`

is `$\phi(x)=g^x \mod p$`

.

- 0: Order 1 (it's already the identity element)
- 1: Order 6,
`$1+1+1+1+1+1 \mod 6=0$`

- 2: Order 3
`$2+2+2 \mod 6=0$`

- 3: Order 2
`$3+3 \mod 6=0$`

- 4: Order 3
`$4+4+4 \mod 6=0$`

- 5: Order 6
`$5+5+5+5+5+5 \mod 6=0$`

I don't quite get this question. I think I need *more* information about
`$G$`

in order to answer it? Otherwise all I can say about `$\langle x \rangle$`

is that it contains 2015 elements (or alternatively infinitely many).

The joke here is that a group can only have a proper subgroup isomorphic to itself if the group is infinitely big. Hence, the person's love for their partner is infinite.

Sweet.

If we allow **Fact 1.4.7** to be given, then this is easy
to prove: If `$\text{ord } g$`

must divide `$|G|$`

then
`$g^{|G|}=g^{\frac{|G|}{\text{ord }g} \cdot \text{ord }g}=1_G^{\frac{|G|}{\text{ord }g}}=1_G$`

.

However, if we can't assume **Fact 1.4.7** then we haven't made our
job easier.

Assume `$\text{ord } g$`

does not divide `$|G|$`

. Then it is either the
case that (1) `$\text{ord } g<|G|$`

or (2) `$\text{ord } g>|G|$`

.

- Dunno?
- In this case, by the pigeonhole principle, there must be some
`$i, j \in ℕ$`

so that`$g^i=g^j$`

, with`$i<j<\text{ord } g$`

. But then`$g^j \cdot g^{\text{ ord} g-j}=1$`

, but then also`$g^i \cdot g^{\text{ ord} g-j} \not =1$`

, even though they are the same operation. This can't be the case, so we exclude`$\text{ord } g>|G|$`

.

Let the isomorphism `$\phi$`

be as follows