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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))$?

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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

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.

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