Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $N$ be a normal $p$-subgroup of finite centerless group $G$ such that $G/N\cong A_{5}$. Is it possible $G/N\cong \leq $Aut$(Z(N))$?

share|improve this question
    
Background, origin of problem, ideas, insights, self work...? –  DonAntonio Oct 22 '12 at 17:37
    
Is this really the question you want? Every finite group is a subgroup of Aut(Z(N)) for some p-group N. –  Jack Schmidt Oct 22 '12 at 17:42
    
What does $\cong\le$ mean? –  Derek Holt Oct 22 '12 at 18:00
    
@DerekHolt "isomorphic to a subgroup" perhaps? –  Alexander Gruber Oct 22 '12 at 21:37
    
@DerekHolt: Isomorphic to a subgroup. This question related to my former question. –  Ros Oct 23 '12 at 2:54

1 Answer 1

Let $p$ be a prime number and $Q$ be a finite group with no normal $p$-subgroups.

Define $N$ to be the direct sum of the simple $\mathbb{Z}/p\mathbb{Z}[Q]$-modules other than the trivial module. Then $Q$ acts faithfully on $N$, and no nonzero element of $N$ is centralized by all elements of $Q$. Notice $Z(N)=N$. Let $G$ be the semi-direct product of $Q$ acting on $N$.

Then $G$ is a finite centerless group with $G/N \cong Q$, and the map $gN \mapsto ( n \mapsto g^{-1} n g)$ is an injective homomorphism from $G/N$ into $\operatorname{Aut}(Z(N))$.

If $Q$ has normal $p$-subgroups, then $G$ is no-longer centerless, and we may need to choose a more complicated $N$ (still abelian, but possibly including the trivial module, and possibly not semi-simple) in order to embed $G/N$ in $\operatorname{Aut}(Z(N))$.

Of course for $A_5$ we need not worry about normal $p$-subgroups, and we can be pretty explicit about $N$. However, I don't understand what could motivate you to ask this question and suspect you meant to ask something else.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.