Mathematics - 2020-02-18

1:47 PM

If $a$ and $b$ are group elements, is it true that $\langle a,b \rangle / \langle a \rangle \cap \langle b \rangle \cong \langle a \rangle \times \langle b \rangle$?

If $a=(1 2)$ and $b=(2 3)$ in $S_3$, then $\langle a,b\rangle=S_3$ and $\langle a\rangle\cap\langle b\rangle=\{()\}$, but $\langle a\rangle\times\langle b\rangle\cong V_4$

Ah, okay. If they commute, then everything should work out, right?

Nah. Actually, yesterdays counterexample also works for this.

oh shoot...So, is it still true that if $a$ and $b$ commute, and $\langle a \rangle \cap \langle b \rangle = \emptyset$, then $\langle a,b \rangle \cong \langle a \rangle \times \langle b \rangle$?

well, the intersection should contain the identity, but yes, that's true

1:55 PM

Whoops, you're right!

So, the isomorphism is presumably $a^i b^j \mapsto (a^i, b^j)$, right? I can see why commutativity is needed, but when do I need that $\langle a \rangle \cap \langle b \rangle = \{1\}$ to prove this map is an isomorphism? I seem to be able to prove that it is both injective and surjective without the assumption...

since $a,b$ commute, every element of $\langle a,b\rangle$ has the form $a^nb^m$ for $n,m\in\mathbb{Z}$, so you can map $\langle a,b\rangle\rightarrow\langle a\rangle\times\langle b\rangle,a^nb^m\mapsto(a^n,b^m)$. This map need not be well-defined, but it is in case $\langle a\rangle\cap\langle b\rangle=\{e\}$. The bijectivity is obvious and the homomorphy follows from commutativity.

Ahh, I see!

try writing out what it means for this to be well-defined and you will see where the hypothesis is needed

2:38 PM

How do you define amalgamated products of groups?

2:50 PM

What's the difference between an amalgamated product and an amalgamated free product?

2 hours later…

Peter

4:24 PM

Anyone wants to factor the 98 digit composite (5^157+4^157+2^157)/2159398397231 ?

Mike Miller

started running a program because i'm curious if it'll finish in a couple hours

Mike Miller

5:19 PM

closed the tab by mistake

so i will not do any more factoring

Mats Granvik

5:41 PM

knotfinite

Peter

@MikeMiller Never mind, I now have the factors.

FF 98 (show) (5^157+4^157+2^157)/2159398397231<98> = 516299365290790543149032460574759294537<39> · 4909715362...23<59>

Ted Shifrin

6:39 PM

@MikeMiller LOL.

2 hours later…

Balarka Sen

9:02 PM

Hey any graph theory dude here

This definitely works right? I was taught some fucked up proof of Hall's marriage theorem by induction but I think you can give a direct algorithmic proof

Isn't that the same thing as the induction proof though ?

Balarka Sen

Maybe man, I couldn't follow the inductive proof

Induction makes everything so non-transparent

Also Hall's theorem is equivalent to Konig's theorem which can be proved, like every minmax result ever, using duality in linear optimization.

I don't know why I have to learn these crazy proofs.

9:28 PM

At some point in your proof you show that if $a\notin M$, then $|N(a)|\ge 2$

You can then argue that if you remove $\{b_0, a_1\}$ from $G$ you still have the same problem but with a graph with 2 less vertices

Do the same over and over until the maximum matching is empty, which is absurd because $|N(a)|\ge 1$ so you can add one edge

That's essentially your proof

Balarka Sen

Hmm, I'll think about it carefully tomorrow when I am more awake.

Also I was a bit glib with what my rematching process was.

I wanted to check the following: Assume $b_k$ is unmatched. If $b_k$ is adjacent to $a_0$ we are done. If $b_k$ is adjacent to $a_1$, delete $(a_1, b_0)$ add back $(a_0, b_0)$ and $(a_1, b_k)$ so done. If $b_k$ is adjacent to $a_2$, delete $(a_1, b_0)$ and $(a_2, b_1)$, and add back $(a_0, b_0), (a_1, b_1)$ and $(a_2, b_k)$. So on.

This always gives a new matching which is one better than the existing maximal matching

Wait I think I said something stupid

Hello guys! If I have a recurrence relation with independent term $f(n)=4(1-n)2^n$, then what would be the proposal for the particular solution?

What does that mean ? @manooooh

I tried this: Since $f(n)=4\cdot2^n-4n2^n$, then $a_n^p=A\cdot2^n+(Bn+C)E\cdot2^n$, but I think it is not correct

9:41 PM

What's the recurrence relation ?

@Astyx it is $a_{n+1}-4a_n=4(1-n)2^n$

Yeah my induction step is wrong you can't remove those two points like that @BalarkaSen

10:00 PM

Look for $a_n = 4^{n-1}a_1 + \sum_{k=1}^{n-1}f(k)4^{n-1-k}$ @manooooh

My question is what we have to propose when $f(n)=4(1-n)2^n$, @Astyx

Isn't what said an answer to that ?

$f(n)=4\cdot2^n-4n2^n$, so $a_n^p=A\cdot2^n+(Bn+C)E\cdot2^n$ or $a_n^p=A\cdot2^n+(Bn+C)A\cdot2^n$? For both proposals I can't solve the system of equations for $A,B,C$ (and maybe $E$)

@Astyx no, you wrote $a_n$ but no $a_n^p$ (the particular solution with indeterminate coefficients)

Also I have no idea how you get that $a_n$, I have never seen it before

Well $a_2 = 4a_1 + f(1)$, so $a_3 = 4a_2 + f(2) = 4^2 a_1 + 4f(1) + f(2)$

Sorry, I have never work like that

Q: Book recommendation for Axiomatic three-dimensional geometry and analytic geometry

10:06 PM

so $a_4 = 4^3 a_1 + 4^2f(1) + 4f(2) + f(3)$ and so on

If you want a specific solution take $a_1 = 0$ for instance

And you have a particular solution to the recurrence relation which is $\sum_{k=1}^{n-1}f(k) 4^{n-1-k}$

Hi @TedShifrin

Akiva Weinberger

You can consistently orient the circle fibers in the Hopf fibration, right?

Ted Shifrin

Hi, @Astyx. Hi, DogAteMy. Yes.

yh05

10:52 PM

I'm reading John Lee's book Axiomatic Geometry and I enjoy it a lot. It includes a detailed treatment of Euclidean plane geometry with rigorous proofs from axioms. I'm looking for books about Euclidean three-dimensional geometry and Cartesian coordinate system (Analytic geometry) with a similar...

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