Mathematics

No, just some of my six eyes opening wider in shock.

robjohn

@MikeMiller ah, spextacles.

@Mike

do you know if such a thing is possible?

Six eyes?

I suppose there are some very strange medical conditions, and some insects have far more.

haha :)

I mean showing that V is not a set

given only 4 axioms

5 I guess

2:27 AM

Ah, I don't know any set theory.

@Mike topologically, the hopf band is the same as (homeomorphic to) an untwisted band, just twisted in how it's embedded right?

neither do I its week one!

is V supposed to refer to something specific that set theorists know about?

what's a hopf band

band with two half-twists (where a mobius band has one half-twist)

2:29 AM

then yeah

homeomorphism type is determined by parity of the number of twists

yeah. Define $V_\alpha = \bigcup_{\beta<\alpha} \mathcal{P}(V_\beta)$ with $V_0=\lbrace\rbrace$

then $V=\bigcup_{\lambda\in\text{ORD}}V_\lambda$

because you can unzip it along a line segment, untwist it fully to something flat, and then zip it back up to get a cuff

oh, i get why it's called the hopf band

@Prototank wow, that's big

Basically I think the proof to show that $V$ isn't a set involves showing that $V=\mathcal{P}(V)$

but I'm not certain

2:32 AM

well, ORD isn't really a set is it?

right but I don't think that we have that

maybe we do?

maybe you can show that if every ordinal embeds (as a set) into your set, then your set is not really a set

but again iunno

I don't know what you mean by "have that." have what? in set theory does it make sense to union over something that isn't a set?

right I think that is a working thing

I'm really sorry anon and Mike for not being clear

I'm in a class and we've only established 5 axioms so far

the one's I've mentioned

no need to apologize, also i didn't notice any lack of clarity

2:34 AM

being "really sorry" should be reserved for situations where you cause trouble

i think that any set that's too big to fail is too big to exist

so when I say "have that", I mean "I don't know for certain that our 5 axioms guarantee that ORD isn't a set"

trying to make a joke about class actions and banks. am failing.

ah

@anon the idea of such a joke made me smile anyway

user147690

3:16 AM

So if $|G|=30=2\times3\times 5$ we get:

$$n_2=1,3,5,15,\quad n_2\equiv 1\pmod{2}$$
$$n_3=1,2,5,10,\quad n_3\equiv 1\pmod{3}$$
$$n_5=1,2,3,6,\quad n_5\equiv 1\pmod{5}$$

And so by the modulo equivalences we eliminate it to:

$$n_2=1,3,5,15,\quad n_2\equiv 1\pmod{2}$$
$$n_3=1,10,\quad n_3\equiv 1\pmod{3}$$
$$n_5=1,6,\quad n_5\equiv 1\pmod{5}$$

Since 3,5-Sylow groups are Cyclic, we have that they intersect trivially

user147690

So then if $n_3=10$ we have $20$ distinct elements, and if $n_5=6$ we have $24$ distinct elements, which means that either $n_3$ or $n_5$ is equal to $1$

yikes

user147690

Oops didn't have chatjax on and editting is over

user147690

Why is there no consideration of 2-Sylow groups?

can you start from the top and say precisely what the question is with no $n_j$s

user147690

3:21 AM

If $|G|=30$, $G$ cannot be simple

there's no consideration of 2-sylow subgroups because there doesn't need to be

since you're trying to get too many elements from groups that aren't normal, $n_2 \geq 3$ is not particularly helpful - it says there are at least 3 extra elements

uhh

from earlier

is it possible to show that $\mathcal{P}(V)\subset V$

user147690

I don't understand