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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 Cayley graph of a group isomorphic to D6 (version 2)
An identical question (just missing the last sentence) was posted yesterday. The first comment contained an answer.
Aug
13
comment $ C_{H}(N) \geq O_{p^{\prime}}(H) $ and $ \overline{H} = H / C_{H}(N) $ is a $ p $-group. Why $ \overline{H} \ltimes N $ is nilpotent?
@Soroush: Are you sure about $p'$ being a prime number? The standard notation in finite group theory is that $p'$ denotes the set of all primes except a fixed prime $p$. Therefore usually $O_{p'}(G)$ is the maximal normal $p'$-subgroup of $G$ where a $p'$-group is a group whose order is not divisible by $p$. If you need a second prime beside $p$, better take $q$.
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?
Aug
2
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.
Aug
2
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.)