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Mar
25
comment Is there a nonabelian finite simple group with Grothendieck ring of multiplicity one?
@DaveWitteMorris : Well spotted. I am not sure at the moment, but I imagine there are others who will know. Notice that the inequality can hold when $G$ is quasisimple. For example, when $G = {\rm SL}(2,5)$ we get $k = 9$ and $b = 6.$
Mar
25
revised Is there a nonabelian finite simple group with Grothendieck ring of multiplicity one?
Expanded
Mar
25
comment Is there a nonabelian finite simple group with Grothendieck ring of multiplicity one?
@JimHumphreys : Oops, sorry, you were right, I misunderstood the sense.
Mar
25
revised Is there a nonabelian finite simple group with Grothendieck ring of multiplicity one?
corrected sense of intended meaning.
Mar
25
comment Is there a nonabelian finite simple group with Grothendieck ring of multiplicity one?
It is not far from optimal in general ( again, ${\rm SL}(2,2^{n})$ is not far from achieving it).
Mar
25
answered Is there a nonabelian finite simple group with Grothendieck ring of multiplicity one?
Mar
24
comment how to find explicitly given component in a regular representation
By the way, the $2$-dimensional irreducible representation of $S_{4}$ is "really" a representation of $S_{3}$, because it has the normal Klein $4$-subgroup of $S_{4}$ in its kernel.
Mar
24
answered how to find explicitly given component in a regular representation
Mar
24
comment how to find explicitly given component in a regular representation
And this is consistent with the fact that the two sided ideal usually associated to the character $\chi$ is $e_{\chi} \mathbb{C}G$, where $e_{\chi}$ is the central idempotent $\frac{\chi(1)}{|G|} \sum_{g \in G} \chi(g^{-1})g.$
Mar
24
comment how to find explicitly given component in a regular representation
It is actually usual to associate the character $\chi$ to the element $\sum_{g \in G} \chi(g^{-1})g \in \mathbb{C}G.$
Mar
23
comment Permutations with fixd points
Proably more suited for MathStackExchange. In any case, you should make it clear whether you mean "with exactly k fixed points",or, "with at most k fixed points".
Mar
22
comment Enumeration of $0-1$ matrices with determinant $1$
A good place to start might be with the upper triangular unipotent matrices $U$ with only $0$s and $1$s above the diagonal, then look at $\sigma u \tau$ for $u \in U$ and $\sigma$ and $\tau$ permutation matrices where the associated permutations have the same sign.
Mar
20
comment quadratic matrix equation
@user35593 : It isn't necessarily true that a complex symmetric matrix is diagonalizable: for example, when $c$ is real and non-zero, the matrix $\left(\begin{array}{clcr} ic & c\\c& -ic \end{array}\right)$ has minimum polynomial $x^{2}$ and is not diagonalizable.
Mar
20
comment generating set for symmetric group $S_n$
I suppose it could be that when $b$ replaces a generator, you get a generating set, but it will no longer be independent when $b$ is not a transposition, $3$-cycle, or double transposition.
Mar
19
comment generating set for symmetric group $S_n$
It is unclear to me why ( for an arbitrary irredundant such set of generators), we can't have some non-identity element $b \in \cap_{k=1}^{n-1} H_{k}$, where $H_{i} = \langle a_{j}: j \neq i \rangle$ for $1 \leq i \leq n-1$, even given Whiston's theorem.
Mar
19
comment generating set for symmetric group $S_n$
I presume you mean an irredundant generating set of size $n-1$ ( ie $n-1$ elements such that no proper subset generates, but the whole set does). If you don't insist on size $n-1$, then there are examples: eg $S_{7} = \langle (12),(1234567) \rangle$, but $S_{7}$ is not generated by $(12)$ and any $5$-cycle. I am not sure how many irredundant generating sets for $S_{n}$ have size exactly $n-1$ -maybe relatively few.
Mar
17
comment Maximal abelian subgroup of general linear groups
Yes, I am sure there is more inductive information available in the unipotent case, but I am also sure that there are people more expert than I am in this particular area.
Mar
17
answered Maximal abelian subgroup of general linear groups
Mar
17
comment Which kind of subsets of primes one needs to generate a positive ratio of the natural numbers?
@RichardStanley : I think there may be typo in your comment??
Mar
16
comment Suppose that $G$ is a subgroup of $GL_n(\mathbb C)$ with finite exponent. Then is $G$ a finite group?
Maybe you had meant "thanks for your remark"? I think I misunderstood what you meant.