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2h
comment The free group of a group and the kernel of a canonical morphism
Yes. First, it is generated as a normal subgroup by these elements, call them $f(x,y)$. Then $e_zf(x,y)e_z^{-1}=f(z,x)f(zx,y)f(z,xy)^{-1}$. It follows that the subgroup generated by the $f(x,y)$ is normal in $F_G$ and hence it generates the kernel of $F_G$ as a subgroup.
22h
comment Compactly generated semi-simple Lie groups
An arbitrary connected topological group is generated by any of its non-empty open subsets.
Apr
15
comment Must the powers of some element always grow linearly with respect to a word metric?
PS: $|g^n|=O(\log n)$ follows if $g$ is conjugate to its square.
Apr
14
comment Must the powers of some element always grow linearly with respect to a word metric?
Actually in Distortion functions for subgroups, in Geometric Group Theory Down Under, Proc. of a Special Year in Geometric Group Theory, Canberra, Australia, 1996, Ed. J.Cossey, Walter de Gruyter, Berlin - New York, 1999, 281-291, Olshanskii already constructed a torsion-free 2-generated group in which every element $g$ satisfies $|g^n|=O(\log(n))$. (Reference 56 in math.vanderbilt.edu/~olsh/publ.html)
Apr
14
comment pseudovarieties and profinite group : do * and g() commute?
What is $gV$? What is $g$?
Apr
14
awarded  Nice Answer
Apr
12
comment Can there be a non-trivial epimorphism (of rings) from a field?
I guess that you mean "epimorphism" in the categorical sense (some people use it for "surjective homomorphism", which is stronger since the ring homomorphism $\mathbf{Z}\to\mathbf{Q}$ is a non-surjective epimorphism). An ambiguity is on what you call "ring": associative? commutative? there are people on mathOverflow using various conventions.
Apr
12
answered Must the powers of some element always grow linearly with respect to a word metric?
Apr
10
comment Can an algebraic number on the unit circle have a conjugate with absolute value different from 1?
OK (you said this after your edit :). We have $f_k(t)=\prod_i (t-x_i^k)$, all $x_i$ are algebraic integers and hence so are the coefficients of $f_k$. So my remark is just a (partly) distinct argument, where instead of using basic Galois theory, I use that the ring of invariants of $\mathbf{Z}[t_1,\dots,t_n]$ under the symmetric group $S_n$ is generated (as a ring) by the elementary symmetric polynomials in $t_1\dots,t_n$, and I apply this to coefficients of the polynomial $\prod (t-t_n)$, and then substitute $t_i$ to $x_i$.
Apr
9
revised Coherent subgroups of $F_2 \times F_2$
fixed a typo
Apr
9
comment Can an algebraic number on the unit circle have a conjugate with absolute value different from 1?
I know what Vieta's formulas are: they describe the coefficients of a monic polynomials in terms of elementary symmetric polynomials on the roots (which have integer coefficients). I refer to something else, namely the fact that any symmetric polynomial can be described as a polynomials (with integer coefficients) on the elementary symmetric polynomials.
Apr
9
answered Coherent subgroups of $F_2 \times F_2$
Apr
7
comment finite stabilizers + compact orbit space => proper action?
I don't think the downvote is justified.
Apr
7
comment finite stabilizers + compact orbit space => proper action?
PS: I tried further to get an examples following the last lines; the main issue is not smoothing (which is actually easy), nor ensuring non-properness and compact quotient, but to ensure that the quotient by the discrete subgroup is Hausdorff.
Apr
7
answered finite stabilizers + compact orbit space => proper action?
Apr
7
comment Can an algebraic number on the unit circle have a conjugate with absolute value different from 1?
It has to be said that the coefficients of $f_k$ are symmetric functions of the roots of $f$, and therefore belong to the subring generated by the elementary symmetric polynomials of $x_1\dots x_n$, that is, because $f$ is monic, the subring of $\mathbf{Z}[x_1,\dots,x_n]$ generated by coefficients of $f$. In particular the coefficients of $f_k$ are integers.
Apr
7
comment Algebraic integer with conjugates on the unit circle
Since the product of norms of $\alpha$ over various completions is 1, it has to be a algebraic unit. Then it follows that the cyclic subgroup $C$ generated by $\alpha$ is bounded for every norm. Since $\mathbf{Z}[\alpha^{\pm 1}]$ is discretely embedded as a subring in some finite product of completions, we deduce that $C$ is finite.
Apr
7
comment Solvable Lie algebras: embedded in upper triangular matrices?
PS: for any non-algebraically closed field $K$ there is a non-$K$-triangulable f.d. solvable Lie $K$-algebra: just take a non-$K$-trigonalizable $n\times n$ square matrix and consider the semidirect product $K^n\rtimes K$ defined by this matrix. In characteristic zero a solvable Lie algebra is triangulable iff all eigenvalues of its elements in the adjoint representation are in $K$.
Apr
7
comment Solvable Lie algebras: embedded in upper triangular matrices?
In char. 0 every finite-dimensional solvable Lie $K$-algebra embeds into upper triangular matrices over some finite extension of $K$ (but not always over $K$, for instance the 3-dimensional Lie algebra of the group of isometries of the Euclidean plane is a counterexample); when it can be embedded into upper triangular matrices over $K$, it is often called $K$-triangulable.
Apr
6
comment Products of subgroups of a free group
@BenjaminSteinberg: it's not assumed in the question