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answered  Is there an easy way to tell if all eigenvalues of a unitary or selfadjoint matrix only have eigenvalues of multiplicity two? 
Aug 16 
revised 
Asymptotic density of finite abelian and solvable groups
Amended first sentence 
Aug 16 
comment 
Asymptotic density of finite abelian and solvable groups
OK, thanks, I've reworded my post. 
Aug 16 
answered  Asymptotic density of finite abelian and solvable groups 
Aug 15 
comment 
Special linear groups contained in symplectic groups
Try the book by Kleidman, or papers of Kleidman and Liebeck 
Aug 14 
comment 
Reverse to Chinese Remainder Theorem
I'm not sure what you can expect here, beyond the trivial $x^{\ast} \leq \frac{m}{2}.$ 
Aug 14 
comment 
When is a cubic polynomial a cube?
@Pietro: I thinks he means the (textspeak) shorthand for "laughing out loud", it's not a Mathematical symbol! 
Aug 11 
comment 
A question on conjugacy classes of central involutions in a finite group
In fact, the number of classes of central involutions is the number of $N_{G}(S)$ orbits by conjugation on $\Omega_{1}(Z(S)) \backslash \{1_{G}\},$ where $S$ is a Sylow $2$subgroup and $\Omega_{1}(Z(S))$ is the subgroup of $Z(S)$ generated by its involutions. 
Aug 9 
comment 
Successive Schur covers
For a perfect group, it stops after (at most) one step. In general, I think the sequence can go on indefinitely ( eg, you can take $G_{0}$ to be be a Klein $$group, and $G_{j}$ to be a dihedral $2$group of order $2^{j+2}$ for each $j.$ 
Aug 6 
comment 
A question on conjugacy classes of central involutions in a finite group
I don't know if you really said what you mean, but if there is a unique conjugacy class of involutions in $G$, the statement is true. By the way, I think you need to specify that $a$ and $b$ are distinct involutions. 
Aug 4 
comment 
If $~(c  b) ^ 2 + 3cb = a^3~$ has nonzero integer solutions, then $~(a,c) \gt 1~$ or $~(b,c) \gt 1$?
If $n$ is nice, it has at least one such factorization with $c$ and $b$ coprime, (which is all you need for the answer) but if $n$ is not squarefree, then there are such (rational integer) factorizations with $c$ and $b$ not coprime (eg $7^{3} = 14^{2}+ (7 \times 14) +7^{2}).$ 
Aug 2 
comment 
Existence a finite capable pgroup of class two
Is a capable group the same as a powerful group? 
Jul 27 
revised 
Isolated elements of primary order ($Z^*$theorem revisited)
typos 
Jul 27 
comment 
quasiprimitive nonsolvable groups
There are two cases to consider: $F^{*}(G)$ of order prime to $p,$ and $F^{*}(G)$ of order divisible by $p.$ In the second case you may further ask what happens whether $F(G)$ has order divisible by $p.$ If it does, the situation should be under control. If it doesn't, then I think that some component (quasisimple subnormal subgroup) will fail to have a regular orbit, and that is quite rare. I am not sure what happens if $F^{*}(G)$ has order prime to $p,$ though in that case, the group is not far from $p$solvable. 
Jul 26 
comment 
quasiprimitive nonsolvable groups
Earlier comment deleted:I had not noticed the characteristic restriction. 
Jul 26 
revised 
quasiprimitive nonsolvable groups
typo 
Jul 26 
answered  quasiprimitive nonsolvable groups 
Jul 25 
comment 
Compute an arbitrary decimal place of $\pi$
You might look up to look up the BaileyBorweinPlouffe results about the hexadecimal representation of $\pi.$ 
Jul 25 
comment 
quasiprimitive nonsolvable groups
Your question needs to be more specific, I think. You need to tell us more precisely what you want/need to know 