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 Jul 14 comment How do we identify $\mathfrak{R}$-automorphisms of a group? In the question you linked one has the additionally property that $f$ fixes all subgroups. Does this help in any way? Jul 14 comment Well-defined map from $G/N$ to $G$ that is a homomorphsim? The theorem of Schur-Zassenhaus is a classic result giving a sufficient condition for such a homomorphism to exist: when the order $|N|$ of the normal subgroup $N$ is coprime to its index $|G/N|$. Jul 14 comment Well-defined map from $G/N$ to $G$ that is a homomorphsim? In this construction one can replace the complement $H$ of $N$ by any of its conjugates (but not all possible complements have to be conjugated). Jul 14 comment Well-defined map from $G/N$ to $G$ that is a homomorphsim? Additionally: If you take $N=\{\pm 1\}$ then $G$ does not even have any subgroup isomorphic to $G/N$. Jul 11 comment If $Q = \langle y \rangle X$ for some element $y$, then $\vert N \vert = p$ if $p$ is odd and $\vert N \vert \leq 4$ if $p = 2$. First question: Given such a pair $(Q, N)$ you can always factor out some normal subgroup of $N$ (take $\langle z\rangle$ for an element $z\in Z(Q)\cap N\ne 1$ of order $p$) to get a smaller example $(Q_0, N_0)$ where the order of $N_0$ is $|N|/p$. So a minimal counterexample to the statement has the claimed orders. Jul 10 comment Group $G$ whose center $Z(G)$ is cyclic and with $G/Z(G)$ commutative @DerekHolt: Any chance that you'll give it a second try? The infinite case would interest me. Thanks! Jul 9 comment $G$ non abelian, order $p^3$ ($p$ prime). Suppose that the center is $p^2$, prove that $\exists\ x$ outside of the center, of order p I doubt that you'll find a proof for the existence of an element of order $p$ outside the center without using that $\langle x, Y\rangle$ is abelian for $x\in G$ and $Y$ a subset of the center (which proves your goal directly as you already know). Jul 8 comment Permutation of cosets As $\gamma$ fixes all subgroups, for every element $g$ one gets that $\gamma(g)$ generates $\langle g\rangle$. Jul 2 comment Rotman's exercise 2.8 “$S_n$ cannot be imbedded in $A_{n+1}$” From your comment I conclude that you realized in meanwhile that $A_4$ has an elementary abelian subgroup of order $4$, i.e., two commuting elements of order $2$. Jul 1 comment Are there groups of order $p^4q^2$ which are not semi-direct product? verret mentioned in a comment to a recent answer by Derek Holt to a similar question the existence of a group of order $144 = 2^4\cdot 3^2$ that cannot be written as semi-direct product of any of its proper subgroups. Jun 30 comment Covering groups @lattice: An easier example of a "double cover" is the quaternion group $Q_8$ with $8$ elements. It is a double cover of the elementary abelian group $V_4 = Z_2\times Z_2$ with $4$ elements. If $Q_8$ acts on something, then in general its quotient group $V_4 = Q_8/Z(Q_8)$ doesn't necessarily act on it, as $V_4$ is not a subgroup of $Q_8$. Jun 29 comment On cyclic decomposition of element in $S_n$ The cyclic decomposition of $x$ of prime order $p$ has to be $1^{n-p}p^1$ (This follows from @whacka 's formula "order = lcm(cycle lengths)"). Jun 29 comment Group $G$ whose center $Z(G)$ is cyclic and with $G/Z(G)$ commutative On second thought, taking a maximal abelian subgroup won't work. Instead assuming $G$ finite, I'd reduce to the case $G$ being a $p$-group. $(g, h)\mapsto [g, h]$ induces a non-degenerate skew-symmetric bilinear map $G/Z\times G/Z \to G'$. I'd induct on the order of $G/Z$ by finding a matching element for an element of $G/Z$ of maximal order. Jun 28 comment Group $G$ whose center $Z(G)$ is cyclic and with $G/Z(G)$ commutative A $p$-group $P$ is called special if its center $Z(P)$, derived subgroup $P'$ and Frattini subgroup $\Phi(P)$ are all equal. If this subgroup is additionally cyclic, then $P$ is called extraspecial. As there are extraspecial groups of order $p^n$ for all odd n and primes $p$, you won't be able to finish your proof with $H = G'$. Try looking at a maximal abelian subgroups $A$ of $G$ instead, and see if $H = A/Z(G)$ works. Jun 27 comment A problem on order of a Group. @AlexM: It also follows from $Hg = G\setminus H = gH$. Jun 24 comment How to compute the automorphism group of split metacyclic groups? (continuation of last comment) So $G$ is generated by $x$ and $y$, and $\phi$ centralizes $x$ and maps $y$ to a conjugate $y^z$ for some $z\in Z_p$ (as $Z_k$ is abelian). Hence $\phi$ is conjugation by $z$, an element of $Z_p$, proving the claim. To get $AGL(1, p)\le Aut(G)$ you probably know already that it follows from $G$ being normal in it with trivial centralizer. Jun 24 comment How to compute the automorphism group of split metacyclic groups? I think you can show that the kernel of the homomorphism $Aut(G) \to Aut(Z_p)$ (that exists since $Z_p$ is characteristic in $G$) is $Z_p$: If $\phi\in Aut(G)$ centralizes $Z_p$, i.e., $\phi(x)=x$ for a generator of $Z_p$, then take a generator $y$ of $Z_k$, which acts by multiplication on $Z_p$, let's say $x^y = x^a$ for some $a\in \mathbb{N}$. Now $x^a = \phi(x^a) = \phi(x^y) = \phi(x)^{\phi(y)} = x^{\phi(y)}$, so $\phi(y) = x^b\cdot y$ for some $b\in \mathbb{N}$. But all elements of the form $x^b\cdot y\in G$ are conjugated by an element of $Z_p$. (to be continued) Jun 24 comment On Thompson conjecture What do you mean with "$N(G) = N(A_n)$"? Is $G$ a group that happens to have the same sizes of conjugacy classes as $A_n$ (with or without multiplicity?)? Jun 23 comment What is subgroup lattice of GL$(n,\mathbb F_q)$? What do you need the subgroup lattice for? Jun 23 comment How can you tell if a normal subgroup induces a semidirect product? An example using the Frattini argument you can find in this answer by Derek Holt.