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 2d comment If $G/Z(G)$ is cyclic then $G$ is abelian – what is the point? An application of your application is mathoverflow.net/questions/211159/… Sep 29 comment Special case of Schur-Zassenhaus theorem @MikkoKorhonen: If you look at the elegant proofs of Schur-Zassenhaus for an abelian normal subgroup and of the basic properties of the transfer homomorphism given in the book "The Theory of Finite Groups" by Kurzweil and Stellmacher, you'll see how closely related both are (as Derek already commented). Sep 21 comment Bruhat Decomposition and Iwasawa Decomposition for Finite Groups? For the Bruhat decomposition you need a BN-pair, which in turn gives you a Tits building. Their classification for rank greater 2 show that you are dealing with a group of Lie type, and not a general group. If you are interested in generalizing theory from Lie (type) groups, take a look at "Subgroup complexes" by Peter Webb in Proceedings of Symposia in Pure Mathematics 47, pp 349-365, from 1987. Sep 14 comment If $G=G_1\times G_2$ does it follow that $H\leq G \implies H=H_1\times H_2$ where $H_1\leq G_1$ and $H_2\leq G_2$ With different letters the same question was asked here. Sep 11 comment Intuition of Prime decomposition in Galois extensions of subgroups of a cyclotomic polynomial Let $\omega = \frac{1+\sqrt{3}}{2}$ be a 3rd root of unity, so $\mathbb{Q}(\omega)$ will be one of your subfields. Over $\mathbb{Z}$ you get $\mathbb{Z}[\omega]/(p) = \mathbb{Z}[X]/(p, 1+X+X^2) = \mathbb{F}_p[X]/(1+X+X^2)$, and therefore $p$ is a prime element of $\mathbb{Z}[\omega]$ iff $1+X+X^2$ is irreducible over $\mathbb{F}_p[X]$. Is this really the case for (let's say) $p=7$? Sep 10 comment left regular representation of $Q_8$ What do you mean with "Can any one please explain this solution?"? Where does the solution come from and which steps do you not understand? Sep 10 comment left regular representation of $Q_8$ Two hints: (1) You have to identify the symmetric group on the set $Q_8$ with $S_8$ (which works since $Q_8$ has $8$ elements). (2) If you know two elements $x, y$ of $Q_8$ that generate $Q_8$ (i.e., $\langle x, y\rangle = Q_8$, then their images under your $f$ will generate a subgroup of $S_8$ isomorphic to $Q_8$. Sep 8 comment How To prove Any Change to $v=a\cdot y + b$ maks $y=(a)^{-1}(v-b)$ Uni. random value @Hetebrij: OK, now I see it. The letters got indices towards the end of the question. Yes, without $b=0$ it cannot work. Thanks. Sep 8 comment How To prove Any Change to $v=a\cdot y + b$ maks $y=(a)^{-1}(v-b)$ Uni. random value @Hetebrij: Why is in your long equation calculating the probability for $z=0$ the condition $f(ay+b)=b$ equivalent to $f(ay+b)=1$ and $b=1$ (3rd to 4th line of equation)? How about the case $f(ay+b)=0$ and $b=0$? [I'd assume $a$ to be randomly drawn from ${\mathbb F}_p^\times$ and $b$ from ${\mathbb F_p}$.] Sep 1 comment Condition under which $HK$ is a subgroup @EricAuld: You can find examples under the accordingly named section of the wikipedia page about the Zappa–Szép product, which I linked in my first comment. A simpler example is the symmetric group $S_4$ on four element: It is the (Zappa-–Szép) product of a $2$- and a $3$-Sylow subgroup. Sep 1 comment Condition under which $HK$ is a subgroup Reading your title I thought of the condition $HK=KH$ (try proving that this equation implies that $HK$ is a subgroup!), reading your question reminded me of the Zappa–Szép product, which has the additional property $H\cap K=1$. Aug 31 comment Is there a “unique factorization theorem” for finite groups? All finite solvable groups (and solvable is necessary) have a Sylow basis, this is a set of pairwise permutable Sylow subgroups of G, one for each prime divisor. Sylow bases are unique up to (simultaneous) conjugacy, so there is a kind of unique factorization for solvable groups. But this leaves open the problem of understanding $p$-groups on one side and non-solvable(especially non-abelian simple) groups on the other side. Aug 13 comment A Rédei $p$-group is the union of its maximal subgroups A maximal subgroup of a finite $p$-group is normal and has index $p$. A group with a unique maximal subgroup is cyclic, hence abelian. Now look at your group modulo the intersection of two maximal subgroups. What group do you get? If this group is the union of its maximal subgroups, then you are done, as the preimages of maximal subgroups under surjective homomorphisms are maximal. Aug 5 comment Why does $\operatorname{SL}_2(3)$ only yield even permutations? Only for the field with two elements you have $SL = GL$. What do you get when looking at the action of the full $GL_2(3)$ on the $1$-dimensional subspaces instead? (The exceptional isomorphisms between alternating/symmetric groups and small matrix groups are probably just coincidences without greater pattern, so I wouldn't hope for a good reason here.) Aug 4 comment Is there any neat way to show $\phi$ is a homomorphism? @Vim: Or you could use $S\setminus \rho'(T) = \rho'(S\setminus T)$ somehow. Aug 4 comment Is there any neat way to show $\phi$ is a homomorphism? @Vim: If you like it abstract, you can use $':Sym(\mathcal{P}(S))\to Sym(\mathcal{P}(\mathcal{P}(S)))$ and look what $\rho''(\{\{1, 2\}, \{3, 4\}\})$ is. Aug 4 comment Is there any neat way to show $\phi$ is a homomorphism? @Vim: If you were able to answer my second question, then you know that $':Sym(S)\to Sym({\mathcal P}(S)), \rho \mapsto \rho'$ is a group homomorphism. You should be able to check that all $\rho'$ preserve cardinality (i.e., $|T| = |\rho'(T)|$ for all $T\subseteq S$) and disjointness (i.e., $T\cap U=\emptyset$ implies $\rho'(T)\cap \rho'(U)=\emptyset$ for all $T, U\subseteq S$). Aug 4 comment Is there any neat way to show $\phi$ is a homomorphism? A permutation $\rho$ of a set $S$ induces a permutation $\rho'$ on the power set ${\mathcal P}(S)$ by setting $\rho'(T) := \{\rho(t)\mid t\in T\}$ for $T\subseteq S$. Which permutation does the inverse $\rho^{-1}$ of $\rho$ induce on ${\mathcal P}(S)$? Which permutation does the product $\rho\circ\sigma$ induce on ${\mathcal P}(S)$ if $\sigma$ is another permutation of $S$? Aug 2 comment A question about semidirect product @Joseph: Yes, as $Z_2^2\times Z_3$ is abelian, conjugation is trivial. In $A_4$ an element of order $3$ never commutes with an element of order $2$, so conjugation is definitely not trivial. If you define two semi-direct product with different homomorphisms $K\to Aut(H)$ the resulting semi-direct products might be anyway the same, but you cannot expect them to be the same. Here they are not as one group is abelian and the other one isn't. Aug 2 comment A question about semidirect product @Joseph: Why do you think that for both groups $A_4$ and $Z_2^2\times Z_3$ you would get the same homomorphism?