Can chatgpt accurately calculate expected lengths in Pascal's triangle?

[fresh_42]

I don't understand this at all. The correspondence theorem goes both ways. Every (normal) subgroup of $G/H$ is of the form $A/H$ where $A$ is a (normal) subgroup of $G$ containing $H.$ Since there are no such normal subgroups besides of $G$ containing $H$ besides $G$ and $H$ (by maximality), a normal subgroup $A/H\subset G/H$ must be either trivial or the full group, i.e. $G/H$ is simple.

Not if $A=H$ is the only possibility. Then we get $H/H=1$ and the correspondence theorem is useless.
Why can't all subgroups with $H\subseteq A$ fail to be normal?

You haven't justified why there are no normal subgroups strictly between $H$ and $G.$

Because the semidirect product isn't a direct product. Look at my matrix example, possibly extended by another row at the bottom. I haven't checked, but I think you cannot factor a single copy of $\operatorname{SL}(3).$

 

[Infrared]

Not if $A=H$ is the only possibility.

$A$ can always be either $H$ or $G.$

Then we get $H/H=1$ and the correspondence theorem is useless.

I don't see the relevance of this statement.

Why can't all subgroups with $H\subseteq A$ fail to be normal?

It can certainly happen that $H$ and $G$ are the only possibilities for $A.$ This is exactly when my argument says $G/H$ is simple.

Because the semidirect product isn't a direct product.

I know what a semidirect product is. Why does this mean there aren't normal subgroups between $H$ and $G?$

 

[fresh_42]

I know what a semidirect product is. Why does this mean there aren't normal subgroups between $H$ and $G.$

And I knew that $H\subseteq H.$

Why not? (and I assume you mean proper)

I'm only saying that there are examples where it is the case that the semisimple part is only a subgroup and does not provide normal factors. It is an opinion driven by the analog situation with Lie algebras. This is also a kind of correspondence.

The correspondence principle simply does not apply in the case that there are no normal subgroups in between. So your proof fails in the situation I gave. $G/H$ is not simple and all subgroups $H\subseteq A\subseteq G$ are not normal. Why should this be impossible?

 

[Infrared]

And I knew that $H\subseteq H.$

I think you misunderstood my earlier post. $G$ is always a normal subgroup of $G$ that contains $H.$ I was suggesting that you meant there are no proper normal subgroups (of $G$) that strictly contain $H.$

The correspondence principle simply does not apply in the case that there are no normal subgroups in between.

This is untrue. The correspondence theorem holds for any group quotient.

 

[Office_Shredder]

.
The correspondence principle simply does not apply in the case that there are no normal subgroups in between. So your proof fails in the situation I gave. $G/H$ is not simple and all subgroups $H\subseteq A\subseteq G$ are not normal. Why should this be impossible?

Is your assertion here that there is a subgroup $K\subset G/N$ that cannot be written as $K'/N$ for some $K'\subset G$? I think this is false - $K'=KN$ should work (also the statement of the correspondence theorem is that it is a bijection)

 

[fresh_42]

Is your assertion here that there is a subgroup $K\subset G/N$ that cannot be written as $K'/N$ for some $K'\subset G$? I think this is false - $K'=KN$ should work (also the statement of the correspondence theorem is that it is a bijection)

No, that's not my example. I think of an example where $G/H$ is not simple, e.g. the sum of simple groups, $H$ is maximal solvable and normal, and all subgroups that contain $H$ are not normal.

 

[fresh_42]

This is untrue. The correspondence theorem holds for any group quotient.

I know. But my example only allows $A\in \{G,H\}$ and neither factor is simple. $G/H=A_5^2\, , \,H/H=1.$

 

[Office_Shredder]

No, that's not my example. I think of an example where $G/H$ is not simple, e.g. the sum of simple groups, $H$ is maximal solvable, and all subgroups that contain $H$ are not normal.

If $G/H$ is not simple, then there is $K\subset G/H$ that is normal. By the correspondence theorem, there must exist $K'$ such that $H\subset K'$, $K'$ is normal in $G$, and $K=K'/H$.

What does $H$ being maximal solvable have to do with anything? This just means $K$ is not abelian, which is fine?

 

[Infrared]

I know. But my example only allows $A\in \{G,H\}$ and neither factor is simple. $G/H=A_5^2\, , \,H/H=1.$

Do you accept the following special case of the correspondence theorem?
Proposition: Let $G/H$ be a group quotient and let $K$ be a subgroup of $G/H$. Then $K$ is of the form $A/H$ where $H\subseteq A\subseteq G$. Furthemore, if $K$ is a normal subgroup of $G/H$, then $A$ is a normal subgroup of $G$.

The example you claim to have directly contradicts this because $G/H=A_5^2$ has a normal subgroup $K\cong A_5$ which can't equal any $A/H$ for normal $A$ if your claim that there are no normal $A$ strictly between $H$ and $G$ is correct.

 

[fresh_42]

If $G/H$ is not simple, then there is $K\subset G/H$ that is normal. By the correspondence theorem, there must exist $K'$ such that $H\subset K'$, $K'$ is normal in $G$, and $K=K'/H$.

I don't see why $K'$ is normal in $G$. I was thinking of $G/H\cong A_5^2$ and $A_5\ltimes H$ not normal, subgroup but not normal.

My idea was that the operation is complicated enough so that $A_5^2\ltimes H \ncong A_5\times (A_5\ltimes H)$ and there will be no normal subgroups. It works fine with Lie algebras and I can't see why this would be false for their corresponding matrix groups.

What does $H$ being maximal solvable have to do with anything? This just means $K$ is not abelian, which is fine?

Just to make sure that $H$ is normal and not simple.

Maybe I should have a closer look at the correspondence principle again. In case you are right (K' normal in G) then I have to review my matrix group example again. And even more, figure out where the correspondence between Lie algebras and their groups breaks down. Here I can have sums of simple subalgebras (and thus not simple) which are not ideals. I thought I could pull this situation into their group analogon.

 

[martinbn]

https://proofwiki.org/wiki/Correspondence_Theorem_(Group_Theory)

 

[Office_Shredder]

I don't see why $K'$ is normal in $G$. I was thinking of $G/H\cong A_5^2$ and $A_5\ltimes H$ not normal, subgroup but not normal.

Given $K\subset G/N$, let
$K'=\{kn:\ n\in N,\ kN\in K\}$
This is a group:

First we observe $K'$ also contains all elements of the form $nk$since $N$ is normal, $k^{-1}nk=n'\in N$. Hence $kn'=nk$, and the left side is is contained in $K'$.

In particular this means $n^{-1}k^{-1}\in K'$ since $n^{-1}\in N$ and $k^{-1}N\in K$. So $K'$ contains inverses.

It's also closed under multiplication. Let $x=knk'n'$. Then$knk'n'=k(k'n''k'^{-1})k'n'$ with $n"'=k'^{-1}nk'\in N$. So $x=kk'n''nk$. Similarly $n"n=kn'''k^{-1}$ for some $n'''\in N$. So $x=kk'kn'''\in K'$.

Lastly, this is normal. For any $g\in G$, $(gkg^{-1})N\in K$ by normality of $K$, which means $k=g^{-1}k'g$ with $k'N\in K$. so $gkng^{-1}=k'gng^{-1}$. By normality of $N$ this lies in $K'$

 

[Office_Shredder]

I saw fresh posted a number for 4, but since there's no answer key here (or even a guarantee the problem is easily solved) I feel like we need a bit more.

#5.) .I take a piece of string of length 1, and do the following k times: cut the string into a ratio of 2 to 1 (so a string of length 1 is cut into a string of length 2/3 and a string of length 1/3) I then throw out one of the pieces, leaving myself with a single piece that I can repeat this process on, until I have made k cuts.

What is the expected length of the final piece of the string in terms of k, if
a.) Each time I pick one of the two pieces to throw out randomly with equal probability
b.) Each time I "grab" a random spot on the string and throw that piece out - i.e. I have a 2/3 probability of throwing out the longer pieceAlso, can someone explain what the issue with zorn's lemma is for the maximal normal subgroup?

 

[fresh_42]

I saw fresh posted a number for 4, but since there's no answer key here (or even a guarantee the problem is easily solved) I feel like we need a bit more.

We have $a+b+c=72$ and want to maximize $(\underbrace{a-b}_{=:x})^2+(\underbrace{b-c}_{=:y})^2+(\underbrace{c-a}_{=:z})^2,$ i.e. maximize $x^2+y^2+z^2$ under $x+y+z=0.$ Thus
\begin{align*}
x^2+y^2+z^2=x^2+y^2+(x+y)^2 &\longrightarrow \text{ maximize }\\
x^2+y^2+xy&\longrightarrow \text{ maximize }\\
a^2+b^2+c^2-ac&\longrightarrow \text{ maximize }
\end{align*}
Since all $a,b,c\ge 1$ we get the maximum for $a=c=1$ and $b=72-a-c=70.$ Hence
$$
(a-b)^2+(b-c)^2+(c-a)^2= 2\cdot 69^2 = 9522
$$
For real positive numbers we would get $10368-O(\varepsilon )$ and for any real numbers infinity.

 

[fresh_42]

Also, can someone explain what the issue with zorn's lemma is for the maximal normal subgroup?

The set of proper normal subgroups $S:=\{H\triangleleft G\,,\,\subseteq \}$ is a partially ordered set and not empty, since $\{1\}\in S.$ If $G$ is finite then there is trivially a largest proper normal subgroup. If $G$ is infinite this follows from the AC, resp. from Zorn's lemma as the statement of choice in algebra:

A partially ordered set, in which each chain has an upper limit, contains at least one maximal element.

For those who are interested:
Hewitt, Stromberg Real and Abstract Analysis, GTM 25
has (theorem 3.12 page 14ff.) the proof of equivalence of

• Axiom of Choice
• Tuckey's Lemma
• The Hausdorff Maximality Principle
• Zorn's Lemma
• The Well-Ordering Theorem

 

[Office_Shredder]

The set of proper normal subgroups $S:=\{H\triangleleft G\,,\,\subseteq \}$ is a partially ordered set and not empty, since $\{1\}\in S.$ If $G$ is finite then there is trivially a largest proper normal subgroup. If $G$ is infinite this follows from the AC, resp. from Zorn's lemma as the statement of choice in algebra:

I thought there was a claim that zorn's lemma didn't work, I guess i got confused.

That's a nice solution for 4, thanks.

 

[Infrared]

I thought there was a claim that zorn's lemma didn't work, I guess i got confused.

It doesn't here. The issue is that not every chain of proper normal subgroups has an upper bound, since their union can be the whole group. Contrast with the argument for why every commutative ring has a maximal ideal- an increasing union of proper ideals cannot be the whole ring because none of them contain $1.$ Just to give an example: $(\mathbb{Q},+)$ has no maximal proper (normal) subgroup.

 

[Office_Shredder]

#6.) If you start an infinite random walk on the integers where each step is 50/50 to be one unit left or one unit right, show the expected number of times you return to the origin is infinite.

Bonus question: what about a walk in 2 or more dimensions? Slightly handywavy arguments are acceptable.

 

[Office_Shredder]

#7,). Prove or give a counterexample: Every irreducible complex representation of a compact group is unitary

A lot of words there. Is irreducible necessary? Does the representation need to be continuous? Maybe someone will find out!

 

[martinbn]

#7,). Prove or give a counterexample: Every irreducible complex representation of a compact group is unitary

A lot of words there. Is irreducible necessary? Does the representation need to be continuous? Maybe someone will find out!

This is too well known and easy to find if not known. It is probably too hard for ChatGPT though.

 

[Office_Shredder]

#8) Prove or give a counterexample:
Every ideal in a commutative ring is the intersection of the prime ideals containing it.

 

[fresh_42]

$$12\mathbb{Z}\subsetneq 6\mathbb{Z}=\bigcap_{p|12} p\mathbb{Z}$$

 

[Infrared]

#7,). Prove or give a counterexample: Every irreducible complex representation of a compact group is unitary

A lot of words there. Is irreducible necessary? Does the representation need to be continuous? Maybe someone will find out!

I think representations of topological groups are by definition continuous. If you don't require continuity, this looks very false. For example, consider the group $GL(n,\mathbb{C})$ equipped with the indiscrete topology, which trivially makes it compact. Then the identity map (which is not continuous) $GL(n,\mathbb{C})\to GL(n,\mathbb{C})$ can be considered as a representation, which is certainly not unitary. The righthand side of course has the standard topology.

Assuming continuity, if the group is also Hausdorff, then it possesses a Haar measure $\mu$ and any representation is unitary, using the inner product $\langle u,v\rangle=\frac{1}{\mu(G)}\int_G \langle gu,gv\rangle_0 d\mu(g)$ where $\langle \rangle_0$ is any fixed inner product.

Even if the group $G$ is not Hausdorff, the image of the representation $\rho:G\to GL(V)$ is a compact subgroup $\rho(G)\subset GL(V)$ which is Hausdorff (since $GL(V)$) is. So we just apply the above argument to the group $\rho(G)$ instead, thinking of $\rho(G)\hookrightarrow GL(V)$ as a representation.

I don't see where irreducibility should factor in.

 

[Office_Shredder]

I think representations of topological groups are by definition continuous. If you don't require continuity, this looks very false. For example, consider the group $GL(n,\mathbb{C})$ equipped with the indiscrete topology, which trivially makes it compact. Then the identity map (which is not continuous) $GL(n,\mathbb{C})\to GL(n,\mathbb{C})$ can be considered as a representation, which is certainly not unitary. The righthand side of course has the standard topology.

Nice example.

In the case where the representation map is continuous, I was thinking of a bit of a different argument. Let's suppose we have some representation, and the image is a Hilbert space (otherwise it doesn't make sense to ask if it's unitary). There is a natural norm defined on operators, $||A||=sup_{||x||=1} ||Ax||$. . This is only defined for linear operators that are continuous (in infinite dimensions this could be infinity, but then $A$ it's not continuous)

This norm defines a topology on continuous operators. If this is the topology with respect to which the representation is continuous (which requires the image to lie in the space of continuous operators), then claim, for $A=\rho(g)$ in the image of the representation, $||Ax||=1$ for all unit vectors. For finite dimensions this follows by compactness (there is a vector that maximizes the norm, you can show it's an eigenvalue, hence $A^n$ or $A^{-n}$ has arbitrarily large norm) therefore the representation is unitary. . I'm not actuality sure if it holds for infinite dimensional representations though.

Also, new question.

#9) Prove every lebesgue measurable set of positive measures contains measurable subsets of arbitrarily small positive measure, or provide a counterexample.

 

[Infrared]

there is a vector that maximizes the norm, you can show it's an eigenvalue

This doesn't look true to me: The matrix $A=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$ has only a one dimensional eigenspace $\text{Span}\left(\begin{pmatrix} 1 \\ 0\end{pmatrix}\right)$ but $\text{max}_{||x||=1} ||Ax||$ occurs when $x=\pm \begin{pmatrix} 0 \\ 1 \end{pmatrix}$. This specific matrix can't come from a compact representation because it generates a non-compact subgroup though.

It looks like you're arguing that if $H$ is a Hilbert space, then every continuous representation $\rho:G\to GL(H)$ lands in $U(H)$ when $G$ is compact. Even in the finite-dimensional case, I think this is false. Since $U(n)$ is not a normal subgroup of $GL(n,\mathbb{C})$ pick an invertible matrix $A$ for which $A U(n) A^{-1}\neq U(n)$ and use the representation $U(n)\to GL(n,\mathbb{C})$ given by $U\mapsto AUA^{-1}.$

My argument was to show the weaker claim, which I interpreted as the problem statement, that if $\rho:G\to GL(V)$ is a continuous representation of compact $G$, then there exists an inner product on $V$ for which every $\rho(g)$ is unitary with respect to that inner product (but the inner product should not be set in advance).

 

[Infrared]

9. This seems suspiciously easy but I'll write something down: Let $A$ be a measurable subset of $\mathbb{R}$. For $x\geq 0$ define $f(x)=m(A\cap [-x,x]).$ Note that $f$ is (uniformly) continuous because $|f(x)-f(y)|\leq |x-y|.$ Since $f(0)=0$ and $\lim_{x\to\infty}f(x)=m(A),$ by the intermediate value theorem, $f$ should attain every value in $[0,m(A)),$ so $A\cap [-x,x]\subset A$ can be made to have arbitrarily small measure. The same argument works in $\mathbb{R}^n$ replacing $[-x,x]$ with the closed ball of radius $x.$

 

[Office_Shredder]

This doesn't look true to me: The matrix $A=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$ has only a one dimensional eigenspace $\text{Span}\left(\begin{pmatrix} 1 \\ 0\end{pmatrix}\right)$ but $\text{max}_{||x||=1} ||Ax||$ occurs when $x=\pm \begin{pmatrix} 0 \\ 1 \end{pmatrix}$. This specific matrix can't come from a compact representation because it generates a non-compact subgroup though.

Shoot. I dug into this a little bit, and the maximum value occurs at an eigenvector of $A^*A$. The rest of my post quickly falls apart, whoops.

I like your answer for #9, and it gives a much stronger result than my approach which was to cover the whole space in $\epsilon$ width intervals and observe one intersection must have positive measure by additivity.

#10 feels like it might be a standard result every first year undergrad knows, but it doesn't quite match the ones I thought of when I saw these words z so might even be false.

#10) Prove or disprove: Every uniformly continuous function that is differentiable on a closed interval is necessarily Lipschitz continuous.

I think differentiable on a closed interval requires the one sided derivatives to exist on the endpoints, but if dropping that changes the answer I'm interested in that fact.

 

[Infrared]

#10) Prove or disprove: Every uniformly continuous function that is differentiable on a closed interval is necessarily Lipschitz continuous.

Let $f(x)=x^2\sin(1/x^2)$ for $x\neq 0$ and $f(0)=0,$ let's say defined on $[-1,1].$ We see that $f'(x)=2x\sin(1/x^2)-\frac{2}{x}cos(1/x)$ for $x\neq 0$ and $f'(0)=\lim_{h\to 0} \frac{h^2\sin(1/h^2)}{h}=0.$ As $f$ is continuous on a compact domain, it is uniformly continuous. Since $f'$ is not bounded, but any differentiable Lipschitz function has derivative bounded by its Lipschitz constant, this $f$ cannot be Lipschitz.

 

[Office_Shredder]

#11) given three points on a plane, not collinear, prove there exists a unique circle passing through them.

 

[Office_Shredder]

#12) Prove or disprove: Every graph of 30 vertices has two vertices with the same degree (same number of edges)

Is 30 important here? It actually asked me a question about friends that might have been a mix between a graph theory question and the birthday problem but this is what I reduced it to.