i think i get Teds confusion since $ u$ must be zero when the last variable is zero .Cant translate the problem at one variable.

i would need a function that is zero at the end points of an interval kinda

nvm im confused haha

well, there are not that many harmonic one variable functions.

yes i know. but reflecting a line also does not make it a C^2 function

the point was that harmonic was something crucial that you mentioned late.

not that one variable digressions are interesting.

yes its necessary as you mentioned

to use a mean value property i would need that my $v$ is $C^2$

hi all, given a group $G$, why is the set of elements of $G$ of order $p$ (for some prime $p$) a subgroup of $G$?

what is the order of the identity?

mm, we need more hypotheses? f(x) = -x and g(x) = 1-x have order 2 in the group of bijections from Z to Z. but f(g(x)) = -(1-x) = x-1 has infinite order.

even in finite groups, e.g. S_3, (12)(23) doesn't have order 2.

yes, the claim is false

the lazy counterexample is the universal counterexample given by the presentation $\langle x,y\mid x^p,y^p\rangle$

i was being all socratic and the like

(though it's not entirely trivial this actually is a counterexample)

Z2xZ2 ?

that's odd. i'm reading that the set $\{x \in G : x^p = 1\}$ is a characteristic subgroup of $G$

to copper hat's point, that's order dividing p, not strictly order p

$1^p = 1$ but the order of $1$ is $1$.

i'm full of gems today

still not a subgroup without more hypotheses.

there's probably something about $G$ that you know that we don't

the subgroup generated by the elements of order dividing p would be a characteristic subgroup, i guess. if G is finite abelian this is just the set of elements having order some power of p (including 0).

i think? maybe i'm missing something.

a lot of undercurrents today. sort of a maths cold war

what is G really?

that hits the spot

@leslietownes Are you talking about the G-spot?! ;P

yes, in fact, it's something much stronger known as a verbal subgroup

hold on, my repressed ex-cathlic self can';t handle that

hahahahaha what is happening

verbal subgroups are not only characteristic, but fully characteristic

the possibilities here have suddenly become infinite

also the elements whose p-th power themselves are of course characteristic (fully characteristic even) if they are a subgroup, but they needn't be in general

being abelian is an easy sufficient condition for them constituting a subgroup

@copper.hat ;D

:-)

i'm trying to show that a minimal normal subgroup of a finite solvable group is an elementary abelian $p$-group. to my understanding, this is a group of the form $\{x \in G : x^p = 1\}$ (i.e. all nonidentity elements having order $p$) that is abelian

having order that divides $p$...

i love ambiguities in poetry, not so much in maths

i'm confused then, how can such a group have an element that divides $p$ if $p$ is prime? (other than the identity)

aahh i remember that excercise

its a classic

every group has an identity. the order of the identity is one

it seems like the order of implications is the wrong way round here

hmm, and i'm not even drinking yet

being an elementary abelian p-group means that every element has order dividing p, not that it is the set of all elements having order dividing p

i can't delete something after 42 milliseconds

where's a mod when you need one...

@Thorgott I assume you mean "order divided by p"

no, this is an instance where I actually meant it that way

so order 1 or p?

ok I should have read before bothering

ye, either order p or the identity, coppers favorite element

then perhaps i'm looking at the wrong subgroup (let's call it S)? i know i need to use the minimality of a minimum normal subgroup $M$ to show that $M = 1$ or $M = S$. which thus calls for showing that $S$ is normal in $M$

yes take the commutator subgroup of S

copper's already an element

fades away shamefully

i'm having an identity crisis.

i used the commutator and minimality to show that $M$ was abelian. but what about the elementary $p$-group part?

@Jam did you resolve your harmonic problem?

nope im thinking if N normal in H and H normal in G is N normal in G?

gosh, i'm reeling in the points today. looking forward to my mse swag private jet. i think i will go for the global 7500 so i can back to ireland without stopping.

how would i show it is C^2? copper?

the condition $u(x',0) = 0$ is a very strong condition

nothing im stuck :D

i can show it is continuous

it means ${\partial^2 u(x',0) \over \partial x_k^2} = 0$ for $k=1,..,n-1$ and so for $n$.

so i need to prove all partials at x_n=0 are zero

ohh ok

the only points of contention are the behaviour at $(x',0)$.

so i can just say all partials x_n=0 are zero since the function is constant there

and hence c^2

and for all other points is C^2 cause u is C^2

so the only step i need to prove is $u(x',0)=0 => $ all partials are zero

well, since things are $C^2$ on either side of the boundary you just need to show that the (1st & 2nd) partials of $v$ are continuous at that boundary. then you can conclude that $v$ is $C^2$.

then since $u,v$ are harmonic and agree on an open set you can conclude more..

thanks copper. Ill try to prove that being constant results the partials are zero tommorow

@copper.hat im beat.

good luck!

ill just tkae the limit i guess and the numerator will always be zero

hold on, if $u(x',0) = 0$ then you can evaluate the partials wrt to $x_k, k=0,..,n-1$ easiliy.

$\partial \frac{u}{x_i}=\frac{u(...,x_i+h,...,0)}{h}$ as h goes to zero

what is the derivative of a constant function

zero ofc

and since $u$ is harmonic what can you conclude about ${ \partial^2 u(x',0) \over \partial x_n^2}$?

ohh right

nice

but saying u is constant so the derivative is zero is fine?

is like saying f(x)=x is 1 at 1 so the derivative at 1 is zero

ohh nevermind

the numerator of the limit of the partial derivative is zero always

obviously

for every $h$

yeye got it thanks

you just need $u$ and the 1st & 2nd (continuous) partials to match at the boundary.

damnn it was easy . I just dont have the mind for these kind of math :P

sometimes you just need to step back a little :-).

im not used to mult calculus and pde's not my fav course

well i got a lill better today ^^

that's the way life goes :-)