Reputation
5,701
Next privilege 10,000 Rep.
Access moderator tools
Badges
2 12 27
Newest
 Electorate
Impact
~55k people reached

Feb
5
comment Regarding part of proof of proposition: Any topological group $(G, \tau)$ which is a $T_1$-space is also a Hausdorff space.
In a topological group, the multiplication map $g\mapsto xg$ is a homeomorphism for any fixed $x\in G$. So the open set $V$ maps to open sets $xV$ or $yV$ under these homeomorphisms.
Jan
25
comment Linear Algebraic Group acting on the co-ordinate ring
@JagdeepSingh If $V$ is an affine variety, ususally $C[V]$ is identified as functions from $V\to k$, $k$ being the underlying field. So $f$ is a polynomial, but if you actually plug in $x^{-1}y\in G$, you end up with an element of $k$. Maybe more clearly, if you denote $g=x_2\cdot f\in C[G]$, then $(x_1\cdot(x_2\cdot f))(y)=(x_1\cdot g)(y)=g(x_1^{-1}y)=(x_2\cdot f)(x_1^{-1}y)=\cdots$,
Jan
21
answered Linear Algebraic Group acting on the co-ordinate ring
Nov
3
answered Suppose $\lambda$ is an eigenvalue of $T$. Prove $\lambda = 2$ or $3$ or $4$
Sep
21
comment Trying to prove that a function is bilinear.
Since $B\colon V\to V^\ast$, and since $w\in V$, $B(w)\in V^\ast$, which means $B(w)$ is a linear functional on $V$, that is $B(w)\colon V\to K$. So $B(w)$ is not in $V$, but is a linear transformation sending $V$ to the scalar field. So you can apply $B(w)$ to elements $v\in V$ as you would with any linear transformation with domain $V$.
Sep
20
comment Number of conjugates of an $m$-cycle $\sigma$ in $S_n$
Two permutations are conjugate iff they have the same cycle type, so you're just counting the number of $m$-cycles in $S_n$. If you write out an $m$-cycle, there are $n$ choices for the first spot, $n-1$ for the second etc. But then you have to divide by $m$ to account for cyclic permutations, since it doesn't matter what the first entry of the cycle is, just the order in which the entries are written, up to cyclic permutation, e.g. $(213)=(321)=(132)$.
Sep
20
comment There is only one possible class equation of G
@Dude I know, I'm just curious because non-abelian groups of order $15$ don't exist.
Sep
20
comment There is only one possible class equation of G
Where did you find this question? Any group of order $15$ is abelian, cyclic even.
Sep
10
comment The definition of a torus
If it's popping up in group theory/AG, maybe they mean algebraic torus?
Sep
9
revised Problem from “The Theory of Finite Groups” by Kurzweil and Stellmacher
edited title
Sep
9
answered Problem from “The Theory of Finite Groups” by Kurzweil and Stellmacher
Sep
8
comment find the factor groups of the p group - G and prove that G is solvable , where $|G|= p^a$ , p is prime
$p$-groups are solvable: math.stackexchange.com/questions/333977/…
Sep
3
revised Nilpotent ideal and ring homomorphism
added 110 characters in body
Sep
3
answered Nilpotent ideal and ring homomorphism
Sep
2
comment Is any Direct Summand of a Free Module over a PID also Free?
Any submodule of a free module over a PID is free, regardless of whether the free module is finitely generated or not. The proof is not trivial, you can find one here.
Aug
30
revised Why is the arrow ideal $R_Q$ of a finite, connected, acyclic quiver $Q$ equal to the Jacobson radical?
edited body
Aug
30
answered Why is the arrow ideal $R_Q$ of a finite, connected, acyclic quiver $Q$ equal to the Jacobson radical?
Aug
28
awarded  Electorate
Aug
25
awarded  Enlightened
Aug
25
awarded  Nice Answer