author: niplav, created: 2024-01-26, modified: 2024-10-24, 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?
$(1_G, 1_H)$$(g_1, h_1)^{-1}=(g_1^{-1}, h_1^{-1})$
- (a) Rational numbers with odd denominators (in simplest form), where the operation is addition. (This includes integers, written as
$n/1$, and$0 = 0/1)$.
This is indeed a group.
$+$ is associative.Now, is the $+$ operation closed?
It looks like the denominator must stay odd, but I'm not sure that's necessary.
Assume $(2m+1)a+(2n+1)b=u \cdot k$ and $4mn+2n+2m+1=l \cdot k$. Then
$k$ must be greater than or equal to three.
- (b) The set of rational numbers with denominator at most 2, where the operation is addition.
I assume we're excluding denominator zero.
Then we have identity (0), associativity and the inverse (again the negative). The operation looks pretty closed to me as well.
- (c) The set of rational numbers with denominator at most 2, where the operation is multiplication.
This set is not a group, because with the identity element $1$ the
number $3$ doesn't have an inverse.
- (d) The set of nonnegative integers, where the operation is addition.
This set is also not a group because it doesn't have the inverse for,
e.g., the number $1$.
$x \mapsto gx$ is an injection: Assume there is a $y$ so that no $x$ so that $gx=y$. Then let $g^{-1}y=x'$ (and ignore the suggestive naming). But then $gg^{-1}y=gx'$ and therefore $y=gx'$. So such a $y$ can't exist.$x \mapsto gx$ is an surjection: Assume $x \not =x'$. Assume also $gx=y=gx'$. Then $gx=gx'$. But then $g^{-1}g=g^{-1}gx'$, so $x=x'$.A thing that tripped me up was that I then tried to prove that right multiplication isn't a bijection—only to give up in confusion and later find out that it is also a bijection. So much for suggestive questions.
Let $g$ be the primite root modulo $p$. Then the isomorphism between
$ℤ/(p-1)ℤ \cong (ℤ/pℤ)^{\times}$ is $\phi(x)=g^x \mod p$.
$1+1+1+1+1+1 \mod 6=0$$2+2+2 \mod 6=0$$3+3 \mod 6=0$$4+4+4 \mod 6=0$$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) $\text{ord } g>|G|$.
$i, j \in ℕ$ 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: $\phi(1)=\{1, 2, 3\},
\phi(s)=\{1, 3, 2\}, \phi(r)=\{3, 1, 2\}, \phi(r^2)=\{2, 3, 1\}, \phi(rs)=\{2,
1, 3\}, \phi(r^2s)=\{3, 2, 1\}$.
I could go through the pairs of elements of $D_6$ individually, but
that seems not smart.