user736948

@loch actually this should be true even if the group is non-abelian. Or am I missing something?

Oh I guess I'm thinking it's abelian...

@loch xyxy=(xy)^2=x^2y^2=e

@But xyxy=e no?

What are it's members?

How can the group with presentation <x,y: x^2=y^2=e> be infinite?

How does one check that a solution of the Euler-Lagrange equation is actually an extremal of the functional?

@TobiasKildetoft Thanks! That clarifies things for me.

To be precise... the generators are the equivalence classes of each letter, no? @TobiasKildetoft

So the canonical inclusion mentioned in my previous post just sends each letter to it's corresponding equivalence class

Oh!

Set of equivalence classes of words with letters in S. Group operation being concatenation.

cheers

The polynomial has exactly four roots, if it has exactly one root in the 1st quadrant then it has exactly one root in the 4th quadrant. So 2 roots must be in quadrants 2 and 43 , but as roots come in conjugate pairs, we must have exactly one root in each...

Oh I get it... being stupid.

"It suffices to show that there is exactly one root in the first quadrant because it is a polynomial with real coefficients, and zeros of polynomials come in conjugate pairs."

In this post why does it suffice to show that the polynomial has exactly one root in the first quadrant? math.stackexchange.com/questions/3027908/…