author: niplav, created: 2024-01-26, modified: 2025-01-30, 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
I'll limit my time to 15 minutes/exercise—not because I don't like chewing on problems (I do!), but because I want to spend most of my chewing on important or unsolved problems.
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: $\phi(1)=\{1, 2, 3\},
\phi(s)=\{1, 3, 2\}, \phi(r)=\{3, 1, 2\}, \phi(r^2)=\{2, 3, 1\},
\phi(sr)=\{3, 2, 1\}, \phi(sr^2)=\{2, 1, 3\}$.
I could go through the pairs of elements of $D_6$ individually, but
that seems not smart, since (even if I leave out the identity), there's
$5^2=25$ different pairs.
Maybe I can just focus on the "composed" operations $r^2, sr, sr^2$?
In that case it's enough to check the following ones:
But I'm not sure that that's actually enough.
As for $D_{24} \not \cong S_4$: Maybe it's
enough to prove that if two groups have two different
multisets of orders, then they
can't be isomorphic.
Proof: Say $G_1, G_2$ have the two "order profiles" $p_1, p_2$
(that is, sorted lists of the orders of all elements in the group),
that necessarily differ in at least one element $x_1, x_2$ so that
$\text{ord } x_1=k≠\text{ord } x_2$.
Then $G_1 \not \cong G_2$, since such a $\phi$ would need
to fulfill $\phi(x_1^{k-1} \circ x_1)=1$, but we know that
$\phi(x_1^{k-1})\circ\phi(x_1)\not=1$.
Now, what are the orders of the elements of $D_{24}$, $S_4$?
Well, my timer has run out, so I'll go to the next exercise.
Let's be not boring and try to write a constructive proof. For that,
given a group of order $p$, which is prime, one can construct an
isomorphism $\phi$ so that $G \cong ℤ/pℤ$.
By Fact 1.4.7 and Lagrange's theorem, we know that every element
needs to have order $p$.
Of course we fix that $\phi(1_G)=0$. Let's fix an element $g_1$
and its inverse $g_1^{-1}$. We now fix that $\phi(g_1)=1,
\phi(g_1^{-1})=p-1$. Similarly, we fix $\phi(g_1^2)=2$, and
$\phi((g_1^2)^{-1})=p-2$, and so on. (My timer runs out)
(Switching to parentheses for the symmetric group.)
Constructed by labeling the corners of a square and then visually rotating it + flipping the values in my mind. It's not stupid if it works :-)
Hm. Can I be extremely wasteful, and set $n=|G|$, and then
do something with that? Like indicating which element is being
"addressed" by swapping that element with the first element? But then
composition doesn't really work. (God I'm writing like a chain of
thought.)
Unit gets mapped to unit, of course.
Maybe a proof by contradiction? Darn timer's up.
Given (a), this is easy: For any permutation, one can construct a matrix
by permuting the rows (or columns, same difference) of the identity
matrix that encodes that permutation. Any of those row-permuted identity
matrices are still in $\text{GL}_n(ℝ)$. So, given Cayley's theorem and
$G$, we find the subgroup of $S \le S_n$ that $G$ is isomorphic to,
and then for each element of $S$ we can construct the isomorphism to
some subgroup of $\text{GL}_n(ℝ)$ by permuting the rows of $I_n$
according to the permutation in $S$.