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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 $G/P$.
Mar
31
comment finding invariant subgroups under all automorphisms
If all subgroups are normal (= invariant under the inner automorphisms), the group $G$ is called Dedekind group. You should be able to use the info in the wikipedia to reduce your problem to the abelian case, which should be doable. I'd guess that only cyclic groups fulfill your criterion.
Mar
30
comment Prove: $G \cong M \times N$ and $G$ is finite $\Rightarrow order(N)$ is not divisible by 5
All elements of $G$ of order $5$ are in $M$. How about Cauchy?