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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.
Jun
12
comment What do you need to perform Karatsuba multiplication?
The formula holds in any (commutative) ring. You can always choose $a=0$, but the choice of $a$ and $b$ given $x$ and $y$ reminds me of division with rest with remainder and therefore Euclidean rings.
Jun
12
comment What do you need to perform Karatsuba multiplication?
Do you want to know when this formula holds (should be in any ring) or when this formula is useful (e.g., for making multiplication faster)?