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1d
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.
1d
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?
1d
comment A question about semidirect product
@Joseph: No. You wrote "G is defined only by H and K" and "we don't need to consider other homomorphisms", which made me think (probably wrongly) that you had the (wrong) idea that given an inner semi-direct product $G=HK$ that there is only one way how to get an outer semi-direct product at all. But this seems not to be your problem (which I do not understand) with semi-direct products. In your example with $n=12$ you get for $A_4$ a non-trivial homomorphism $Z_3\to Aut(Z_2^2)$, but for $Z_2^2\times Z_3$ the trivial homomorphism.
1d
comment A question about semidirect product
In the case of $G=HK$ with $H$ normal, $K$ acts on $H$ by conjugation giving you the "missing" homomorphism $\Phi$.
Jul
30
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 $n$)?
Jul
30
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.
Jul
30
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 $Aut(G\times S) = Aut(G)\times Aut(S)$).
Jul
30
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.
Jul
30
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 $Q_8$.