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 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. Apr 16 comment $Z(G)$ acts on set of conjugacy classes by left multiplication @mesel: Instead of "extracting" you should factor out the center. Apr 16 comment $Z(G)$ acts on set of conjugacy classes by left multiplication @TobiasKildetoft: The conjugacy class of a central element is its singleton.