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 Apr3 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 Apr2 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 $G/P$. Mar31 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. Mar30 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? Mar30 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/… Mar30 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 $h$). Mar30 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 $12$... Mar29 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). Mar29 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. Mar28 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). 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 revised Why is the number of conjugacy classes modulo 16 equal to the order for a finite group of odd order? made title readable Mar26 suggested approved edit on Why is the number of conjugacy classes modulo 16 equal to the order for a finite group of odd order? Mar26 suggested rejected edit on Why is the number of conjugacy classes modulo 16 equal to the order for a finite group of odd order? 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.