Mathematics

Pedro Tamaroff

8:12 PM

@DonLarynx Oh?

@Huy DG?

diffgeo

DargonGall

8:28 PM

@Huy ugh differential forms

Mike Miller

Ugh?

N3buchadnezzar

uuuuuuuuuuuuuuuuuuuuuugggggggggggggghhhhhhhhhh

@MikeMiller I have 150 problemsets to correct, any comforting words?

Mike Miller

No.

Gato

Hi all!

Studentmath

So, another question. If $G$ is finite abelian group, $o(G)=p_1^{n_1}\dots p_k^{n_k}$, $G$ would have the largest element if $G=C_{p_1^{n_1}}\times \dots \times C_{p_k^{n_k}}$?

Pedro Tamaroff

8:43 PM

@Studentmath I don't understand your question.

"Largest element"?

Studentmath

Element of largest order, compared to other $G$'s with same order.

Pedro Tamaroff

Every finite abelian group has an element of largest order.

I still don't understand the "$G$ would have the largest element if..." part.

Ramanewbie

hi @gato !

Studentmath

@Pedro I mean that, say, $o(G)=2^23^2$. I have 4 different abelian groups of such order, correct?

Ah, I get the point.

No, wait, I don't.

Pedro Tamaroff

I think I understand what you mean.

Studentmath

8:47 PM

Then the one with the element of largest order out of all 4 different groups, would be $G=C_{2^2}\times C_{3^2}$.

Pedro Tamaroff

Say $C_p\times C_p$ has order $2$, but the order of its elements is at most $p$.

Studentmath

The others would have an element of largest order, of smaller order than the rest.

Pedro Tamaroff

Yet $C_{p^2}$ has an element of order $p^2$.

Studentmath

Yeah.

Pedro Tamaroff

Yes, this has to do with exponents @Studentmath.

For any abelian group $G$, there is a least $n$ such that $nG=0$.

This is called the exponent of $G$. Some people call any $k$ such that $kG=0$ an exponent. So the exponent is the least integer that is an exponent of $G$.

$G$ is cyclic if and only if $n=G$.

Studentmath

8:48 PM

Oh. That's nice

Pedro Tamaroff

Note that $|G|$ is an exponent of $G$, so $\exp G\leqslant |G|$.

@Studentmath Try to prove it. Tell me if you're stuck.

Studentmath

So a cyclic group has the largest exponent?

Pedro Tamaroff

Among groups of its order, yes.

That's what you observed.

Studentmath

Cheers, will try to do that exercise

Doesn't it come directly from the fact that the order of elements divides the order of the group?

Pedro Tamaroff

@Studentmath Just write down the proof.

You want to see that $G$ must contain an element of order $|G|$.

Show that if $g\in G$ has maximal order, it has order $\exp G$, @Studentmath.

evinda

9:02 PM

@DanielFischer In my lecture notes, it says that we suppoe that we have a solution $x_n=\sum_{i=0}^n a_i 7^i$ of $x_n^2-2 \equiv 0 \mod{7^{n+1}}$ and then we are looking for a $x_{n+1}$ such that $x_{n+1}^2 -2 \equiv 0 \mod{7^{n+2}}$ and $x_{n+1} \equiv x_n \mod{7^{n+1}}$.
After having shown that there is a $a_{n+1}$ such that $x_{n+1} = \sum_{i=0}^{n+1} a_i 7^i$, it says that we want to have solutions for each $n \in \mathbb{N}$ (of $x^2 \equiv a \mod{p^n}$), i.e. for $n \to +\infty$ and $x_{\infty}=\lim_{n \to +\infty} x_n=\sum_{n=0}^{\infty} a_n 7^{n+1}$.

Studentmath

@Pedro wouldn't it be enough to show that $\exp G$ divides $|G|$?

Pedro Tamaroff

@Studentmath Why?

Studentmath

@Pedro because then we have that $|G|$ is an exponent, and $\exp G\le |G|$

Pedro Tamaroff

$|G|$ is always an exponent.

By Lagrange.

Daniel Fischer

@evinda I don't understand your question. You assume that your $x_n$ satisfies $x_n^2 \equiv 2 \pmod{7^{n+1}}$. The other solution of $a^2 \equiv 2 \pmod{7^{n+1}}$ is then $-x_n$, or, if you wish to write it that way, $7^{n+1} - x_n$.

evinda

9:11 PM

@DanielFischer I haven't understood if it is right to assume that the solution of $x_n^2 \equiv 2 \pmod{7^{n+1}}$ is $x_n=\sum_{i=0}^n a_i 7^i$ or if it should be $x_n=\sum_{i=0}^n a_i 7^{i+1}$.. :/

Daniel Fischer

@evinda $\sum_{i=0}^n a_i 7^i$. It must not be divisible by $7$, so you need a $a_07^0$ term.

Studentmath

Oh, I only now got that you mean to prove that $G$ is cyclic iff $|G|=\exp G$....

Eric Gregor

Are simple functions complex-analytic?

i retract the question

evinda

@DanielFischer A ok.. So should it then be $x_{\infty}=\lim_{n \to +\infty} x_n=\sum_{n=0}^{\infty} a_n 7^n$, with $0 \leq a_n \leq 6$ ?

Daniel Fischer

@evinda Yes. I didn't notice the $7^{n+1}$ there before. That's a typo.

evinda

9:20 PM

@DanielFischer A ok :-) Could you maybe also explain me why we don't expect that $\lim_{n \to +\infty} x_n$ is an integer?

Daniel Fischer

@evinda It can't be an integer. For every integer $k$, there is an $m$ such that $7^m > k^2$. Then you can't have $k^2 \equiv 2 \pmod{7^n}$ for any $n \geqslant m$.

Pedro Tamaroff

@Studentmath Yes. =)

evinda

@DanielFischer I am confused now.. Could you explain it further to me?

@DanielFischer Is it also related to the fact that $x^2-2=0$ has no solution in $\mathbb{Z}$ that $x_{\infty}=\sum_{n=0} a_n 7^n$ is not an integer? :/

Daniel Fischer

@evinda The only way that an integer $k$ satisfies $k^2 \equiv s \pmod{7^n}$ for all $n$ is that $k^2 = s$, since for $7^n > k^2$, the representative of the residue class of $k^2$ modulo $7^n$ is $k^2$ itself. Since $2$ is not the square of an integer, that cannot happen for $s = 2$.

@evinda It is exactly that fact.

Transcript for

$Mathematics$ Mathematics