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1d
comment Recognizing action of semidirect product
@BobJohns: Derek's answer assumes $\rtimes$, i.e., the $S_n$ is not normal. I'm sure it's a typo in your source. How would otherwise the direct product of $n$ general linear groups act on $S_n$ (twice the same $n$)?
1d
comment On group with special properties
@A.G $Aut(G\times S)$ contains $Aut(S)$ with $S$ simple. With simple I meant non-cyclic simple (sorry for not stating this), so this is surely not nilpotent.
1d
comment On group with special properties
Take your (nilpotent) group $G$ of order $3^6$, which has a (central) element $x$ fulfilling (ii). For any simple group $S$ the group $G\times S$ fulfills (i) and (ii) as both $S$ and $G$ are characteristic subgroups (and therefore $Aut(G\times S) = Aut(G)\times Aut(S)$).
1d
comment On group with special properties
If you have an example $G$ for (ii) with $Aut(G)$ nilpotent, you could try $G\times A_5$ instead.
1d
comment Recognizing action of semidirect product
Two questions: (1) Do your vector spaces $V_i$ all have the same dimension? (2) Am I correct in assuming that you did intend to write "\rtimes" instead, i.e., the product of the general linear groups and not the symmetric group is normal? (Your special semidirect product happens to be a wreath product. The wikipedia doesn't seem to be a good source in this case, but try to understand what a wreath product of permutation groups is. This should help you understand these representations.)
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 $Hg = G\setminus H = gH$.