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 Mar27 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! Mar27 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. Mar27 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?) Mar26 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/… Mar26 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? Mar24 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. Mar24 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. Mar22 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 $H$? Mar22 comment About first Sylow Theorem proof Try $H_2 = \langle g, H_1\rangle$. Mar20 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". Mar19 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. Mar17 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 . Mar16 comment Group Permutations Proof @JackM: Thanks, the question had already (for my taste too) many answers. Mar16 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.) Mar16 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.)). Mar16 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.) Mar16 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. Mar16 comment Prove: There is a $g \in G$ such that $\forall$ $x \in X: g \circ x \neq x$ Try calculating the order of the set $\{(x,g)\in X\times G\mid g\circ x = x\}$ in two different ways: Once as sum over the elements of $G$ and once as sum over the elements of $X$. Mar16 comment Prove: There is a $g \in G$ such that $\forall$ $x \in X: g \circ x \neq x$ If you knew this lemma, you could use it nicely. I just wanted to know what tools you have at your disposal. Mar16 comment Group Permutations Proof $a^6 = e$ tells you that the order of $a$ divides $6$. $ab = ba^2$ is better written as $b^{-1}ab = a^2$ telling you that $a$ is conjugated to $a^2$ via $b$, which implies that the order of $a$ cannot be divisible by $2$ (conjugated elements have the same order).