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

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.
Mar
27
comment To show number of left cosets equals number of right cosets
@levitt: In the moment you are trying to prove your claim purely set-theoretical ("If the equivalence classes of two equivalence relations on a set have all a fixed cardinality, then the cardinalities of the two sets of equivalence classes (i.e., quotient sets) are both the same."). If you use a tiny bit of group theory, you can give a bijection between the left and the right cosets. (Hint: Do you know a self-inverse bijection on $G$ that reverses orders in products?)
Mar
26
comment Why is the number of conjugacy classes modulo 16 equal to the order for a finite group of odd order?
Assuming Andreas' guess is correct, you can find a proof under drexel28.wordpress.com/2011/04/23/…
Mar
26
comment Are isomorphisms always constructable?
How are groups given? Do you have the description of both given by a finite set of generators and a finite set of relations each plus the information that they are isomorphic, but you have to find an explicit isomorphism?
Mar
24
comment does minimality condition imply normal p-sylow subgroup >
@MikeTeX: Derek and Geoff did not state it explicitly: $O^{p'}(G)$ exists for all finite groups $G$, no matter if it has a unique $p$-Sylow or not. You should find it mentioned in most text books about finite groups like Aschbacher or Kurzweil/Stellmacher. In the latter book it's called the $p'$-residue.
Mar
24
comment which of the following options are true?
Instead of "So all groups of order $p^3$ are not abelian." you surely meant to write "So not all groups of order $p^3$ are abelian." (and - to be picky - the implication "non-abelian of order $p^3$ implies center of order $p$" does not imply the existence of a non-abelian group of order $p^3$). In mathematics preciseness is important.
Mar
22
comment About first Sylow Theorem proof
Sorry, I forgot the index, I meant $g_2$. In case the corrected hint doesn't help, you can ask yourself the following questions: Which properties does the element $g_2$ have? What's its order? Does it normalize $H_1$? Is it contained in $H_1$? What is the order of the subgroup of $G$ generated by $g_2$ and $H$?
Mar
22
comment About first Sylow Theorem proof
Try $H_2 = \langle g, H_1\rangle$.
Mar
20
comment Symplectic group of elementary abelian group.
An accessible introduction to finite symplectic groups is chapter 8 of Donald E. Taylor's "The Geometry of the Classical Groups".
Mar
19
comment Let $G$ a finite group such that $\lvert G \rvert=pm$, with $p$ a prime and $\gcd(p,m)=1$. $G$ has an unique Sylow $p$-subgroup $P$. Prove $P\lhd G$.
"we know that the normaliser P is exactly the centralizer of P" is wrong. Just take $G=S_3$ and $p=3$. Also I don't understand why you write "There's only one, so $|P|=p$.". I'd understand "$|G|=pm$ with $p$ prime not dividing $M$, so $|P|=p$, but the uniqueness of the $p$-Sylow subgroup has nothing to do with its order.
Mar
17
comment How many nonabelian groups up to isomorphism are of the order $p^4q^4$?
Are you sure you really want to find all non-abelian groups of order $p^4q^4$, not just for the abelian ones? If yes, you could get acquainted with their $p$- rsp. $q$-Sylow subgroups by looking at the links given in the comments to these questions math.stackexchange.com/questions/1095008/… and math.stackexchange.com/questions/1162977/groups-of-order-p4 .
Mar
16
comment Group Permutations Proof
@JackM: Thanks, the question had already (for my taste too) many answers.
Mar
16
comment Prove: There is a $g \in G$ such that $\forall$ $x \in X: g \circ x \neq x$
This looks better! (And looking at Peter's last question he does know Burnside's lemma.)
Mar
16
comment Prove: There is a $g \in G$ such that $\forall$ $x \in X: g \circ x \neq x$
Why the upvote? Peter does not know Burnside's lemma (and what's the connection between having no fixed point free element and the fixed points of (which??) $g$ being the whole set? (I see no indication that the $g$ in the exponent is the identity.)).
Mar
16
comment Mapping vector spaces over two different fields?
Fields have a group structure given by addition and, if you take away $0$, one given by multiplication. Looks like you thought only of multiplication. What becomes out of $(a+b)v = av+bv$? (If you ask instead for a ring homomorphism, there will be only injective ones.)
Mar
16
comment Prove: There is a $g \in G$ such that $\forall$ $x \in X: g \circ x \neq x$
Typo: we can conclude that #Gx > 1 is correct.