Reputation
611
Top tag
Next privilege 1,000 Rep.
Create tags
Badges
9 15
Newest
 Yearling
Impact
~5k people reached

2d
comment An Abstract Characterization of $S_5$ using involutions and their centralizers
Clearer hint for (vi): If $|N_G(P)| = 36$ take $H=N_G(P)$, otherwise take $H=C_G(P)$ which has index $2$ in $N_G(P)$ as $N_G(P)/C_G(P) = Z_2$ ($C_1$ contains a nontrivial automorphism of $P$ and by my first comment there are no others).
2d
comment Proving a group is $PSL(2,q)$ with $q>3$ odd.
What's the intended connection between both theorems? Rough idea would be enough (and appreciated a lot).
Apr
23
comment Groups with “few” subgroups
To be pedantic, you will have to replace the induction on $m$ by a double induction on $n$ and $m$...
Apr
23
comment Groups with “few” subgroups
@DerekHolt: As your proof if phrased in the moment, I have problems seeing why all $SP$ have to be different (but I think you can fix that, for example by working in $N/P$): Take $G=S_4, p^r = 8, k=3$. For $i=2$ I pick $P=V_4$, so $N=G$ and $h=3$. As the $s$ (=two) different subgroups of order $3$ I take two different $3$-Sylow subgroups. But then both give the same $SP = A_4$.
Apr
23
comment Groups with “few” subgroups
@coffeemath: Yes, you are right. It should have been "at most".
Apr
23
comment Groups with “few” subgroups
A finite group $G$ of order $n$ such that the number of elements $x$ with $x^d=1$ is less than $d$ for all divisors $d$ of $n$ has to be cyclic. I wonder, if a variation of this topic can lead to a proof that there are at least as many subgroups as divisors.
Apr
22
comment Does A5 have a subgroup of order 6?
If you are able to calculate the normalizer of $\langle(123)\rangle$, then you have just to check if it contains an element of order $2$.
Apr
22
comment Matrix Derivations-Research
I guess the first sentence makes quite some people stop reading your question (the length is also intimidating). "$DT_3(R)$ is the upper triangular matrix with the diagonal being the same element" cannot be correct, as "the" indicates that there is only one. It should be "a". But then reading on, one gets to know that $DT_3(R)$ is meant to be the set of all $3\times3$ matrices whose diagonal elements are all the same. By the way, is $R$ any ring or the ring/field $\mathbb R$ of the real numbers?
Apr
21
comment subsets of $\mathbb{Z}_2^{p}$ up to permutation equivalence
@ThomasAndrews: I read "fixed permutation of the places" as permutation matrices, not all permutations.
Apr
20
comment An Abstract Characterization of $S_5$ using involutions and their centralizers
Hint for (x): Assuming that Jacobson calls "effective action" what otherwise is known as "faithful action", you can try to show that the kernel $K$ of this action is a proper subgroup of $N_G(V)$, that is centralized by an element of order $5$ (look at $Aut(K))$, so cannot contain elements of order $2$. Then lead $K=Z_3$ normal in $G$ to a contradiction.
Apr
20
comment An Abstract Characterization of $S_5$ using involutions and their centralizers
For (vi) you can use $N_G(P)/C_G(P)\le Aut(P) = Z_2$.
Apr
18
comment The relation of determinants between linear transformation.
Almost. You just forgot the "c" in your last equation ;-).
Apr
17
comment Every normal subgroup of $GL_n(K)$ either contains $SL_n(K)$ or is contained in $Z$
I think you can close the gap using the fact that a (nontrivial) normal subgroup of a primitive permutation group (PGL acts 2-transitive on the projective geometry) has to be transitive.
Apr
17
comment Every normal subgroup of $GL_n(K)$ either contains $SL_n(K)$ or is contained in $Z$
You are right, there is still a gap.
Apr
17
comment Every normal subgroup of $GL_n(K)$ either contains $SL_n(K)$ or is contained in $Z$
Your statement is equivalent to all (nontrivial) normal subgroups of $PGL_n(K)$ containing $PSL_n(K)$. For the exceptions mentioned in my first comment see en.wikipedia.org/wiki/Projective_linear_group#Finite_fields
Apr
17
comment Every normal subgroup of $GL_n(K)$ either contains $SL_n(K)$ or is contained in $Z$
Are you looking for a proof that $PSL_n(K)$ is simple for $n\ge 2$ (with few exceptions) or for a proof why simplicity implies the statement?
Apr
17
comment The relation of determinants between linear transformation.
Hint: Look at the matrices of $L^{-1}\cdot L_c$ and $L^{-1}\cdot L$ with respect to the basis $(v_i)_i$, and use that the determinant is multiplicative.
Apr
17
comment Equivalence of right and left cosets of two different subgroups.
A solution would be: The image of a generator of $Z_5$ has order $5$ in $S_5$, hence is a $5$-cycle. All $5$-cycles in $S_5$ are conjugate. If two elements are conjugate, then so are the subgroups they generated (the converse doesn't hold). For subsets $A, B$ of a group $G$ and an element $x\in G$ the equality $xA = Bx$ is equivalent to $A = x^{-1}Bx$, i.e., to $A$ being conjugated to $B$ via $x$. Now fill in the details...
Apr
16
comment $Z(G)$ acts on set of conjugacy classes by left multiplication
@mesel: The sorry referred to your question. I don't expect any interesting answer.
Apr
16
comment $Z(G)$ acts on set of conjugacy classes by left multiplication
@mesel: The induced action would be trivial. The preimages of the conjugacy classes of $G/Z(G)$ are the orbits. On each orbit $Z(G)$ acts regularly. I guess there is nothing interesting to gain here. Sorry.