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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.
Apr
16
comment $Z(G)$ acts on set of conjugacy classes by left multiplication
For the orbits I'd try to look at the conjugacy classes of $G/Z(G)$.
Apr
14
comment Various Intersections of Sylow p-subgroups.
In the alternating group $A_7$ the $3$-Sylow subgroups $S_1=S_3=\langle (123), (456)\rangle$, $S_2=\langle (123), (457)\rangle$ and $S_4=\langle (124), (356)\rangle$ answer 1). For 2) the examples would become biggish, but should exist.
Apr
14
comment Various Intersections of Sylow p-subgroups.
You should be able to find counterexamples for all claims by choosing the right groups $G$ and $H$ and looking at $G\times H$. The Sylow subgroups of $G\times H$ are the direct products of the Sylows of $G$ with those of $H$, so just pick the latter ones for your needs. For 1) and 2b) you can take $p=2$ and $G=H=S_3$. For 2a) $p=2$, $G=Z_5\rtimes Aut(Z_5)$ and $H=A_5$ (or another group with 2-Sylows isomorphic to $Z_2\times Z_2$ and two 2-Sylows intersecting trivially).
Apr
9
comment Subgroups and Direct Products of $\pi$-closed groups are also $\pi$-closed.
@Stefan: You're right. Sorry. I was going into the wrong direction. I should have asked about the image $X = p_1(O_\pi(G\times H))$ of $O_\pi(G\times H)$ under the projection $p_1 : G\times H\to G, (g, h) \mapsto g$ to the first coordinate. What do you know about $X\le G$?
Apr
9
comment Subgroups and Direct Products of $\pi$-closed groups are also $\pi$-closed.
@Stefan: What is the relation between $O_\pi(G)$ and $O_\pi(G\times H)\cap G$ (considering $G$ as subgroup of $G\times H$)?
Apr
8
comment Subgroups and Direct Products of $\pi$-closed groups are also $\pi$-closed.
A finite group is $\pi$-closed iff the set of all $\pi$-elements is a subgroup.
Apr
6
comment If subgroups are preserved under preimages, is it necessarily a homomorphism?
@goblin: If you look just at subgroups not elements, you cannot distinguish the generators of cyclic subgroups.
Apr
3
comment Proving that a group of order $pqr$ (with conditions on those primes) is abelian.
$1$ (which is the case if both elements commute). Now if $y$ is an element of $P$, all conjugates are in $P$ as $P$ is normal. OK, maybe I should not have called the prime $p$, but $r$ instead. Then you see, as $p\ne 1\pmod r$, that $P$ contains another element commuting with $x$ besides the obvious element $1$. But as $P$ is cyclic of prime order, this element generates $P$ and therefor all elements must commute with $x$. That's how you use these strange conditions that "everything" is not equal $1$ modulo "something else".
Apr
3
comment Proving that a group of order $pqr$ (with conditions on those primes) is abelian.
Given two nontrivial elements $x$ and $y$ of a group $G$, assume that $x$ has prime order $p$. Then there are two possibilities: Either $x$ and $y$ commute, i.e., $xy=yx$, or $y^x := x^{-1}yx \ne y$ (some define $y^x = xyx^{-1}$ instead) is a conjugate of $y$ different from $y$, and so are $y^{x^2}, y^{x^3}, \dots y^{x^{p-1}}$ all different elements. As $x$ has order $p$, $x^p=1$ and so $y^{x^p}=y$. So we have $p$ elements which are all conjugates to each other via $X:=\langle x\rangle$. The technical term is that $X$ acts via conjugation on $G$. Each orbit has length either $p$ or
Apr
2
comment Proving that a group of order $pqr$ (with conditions on those primes) is abelian.
Do you know Sylow's theorem(s)? If yes, prove that there is a unique (hence normal) $p$-Sylow subgroup $P$. First show that $P$ is central, then look at $G/P$.