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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?
Mar
30
comment Is $A_{4}\times Z_2\simeq \langle g,h \mid g^{12},h^2,{gh}^{12}, gh=hg\rangle$?
@Joseph: If you want to know the automorphism group of $Z_4\times Z_6$ then you can take a look at math.stackexchange.com/questions/102895/…
Mar
30
comment Is $A_{4}\times Z_2\simeq \langle g,h \mid g^{12},h^2,{gh}^{12}, gh=hg\rangle$?
According to the theorem in a link you had in an earlier version of your question the statement about the automorphism group of $A_4\times Z_2$ looked correct. But as $G$ is some other group, you first have to identify it (from the presentation I see only an automorphism exchanging $g$ and $gh$ fixing $h$).
Mar
30
comment Is $A_{4}\times Z_2\simeq \langle g,h \mid g^{12},h^2,{gh}^{12}, gh=hg\rangle$?
If you map $g$ to a generator of the cyclic group $Z_{12}$ with $12$ elements, and $h$ to the identity, you see that $Z_{12}$ is a quotient of the group given by the presentation. $A_4\times Z_2$ does not have an element of order $12$...
Mar
29
comment About conjugacy in $A_n$
As the size of the conjugacy class of an element $x$ equals the index of its centralizer $C_G(x)$ in $G$, you have to determine if the given $x\in A_n$ is centralized by an element of $S_n\setminus A_n$ or not (that's what Mark is doing in his answer without stating this fact explicitly).
Mar
29
comment About conjugacy in $A_n$
@Joong: For $n>6$ two 5-cycles are always conjugated, as thanks to their two fixed points you are in the second case of Mark's analysis: both have two cycles of the same odd length 1. Mark mentioned this also at the end of his second last paragraph.
Mar
28
comment To show number of left cosets equals number of right cosets
@levitt: In David's answer you can see the missing part of your proof (your map was not "well-defined", and only for normal subgroups you can prove that). Knowing the trick, one can also prove the statement more direct by defining $X^{-1} = \{x^{-1} | x\in X\}$ for subsets $X$ of $G$. Restricting this function (which is not depending on any representative/choice, hence well-defined) to the left cosets G/H you have to show (1) the image of a left coset is a right coset and (2) for every right coset there is a left coset mapped onto it. Try this proof (and look at the example $S_3$ from chat).
Mar
27
comment To show number of left cosets equals number of right cosets
@TimRaczkowski: How do you think the current approach can be made into a working proof? (see my comments to the questions) If you don't want to solve levitt's homework now, please post it in a week. Thanks!
Mar
27
comment To show number of left cosets equals number of right cosets
@levitt: The problem with the statement of my last comment in the quotes is that it not even correct. You can partition a countable set into just one countable set or also into countably many countable sets. The first quotient set has one element, the other countably many. So I don't see how your current argument is leading towards a proof.