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Let $N$ be a normal subgroup of group $G$. There exists a prime $p$ such that $G/O^p(N)$ is simple group. Is there any element in the center of group $G$?

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May I ask what is $O^p(N)$. I don't know this notation. Thanks. –  Babak S. Oct 19 '12 at 15:53
    
@BabakSorouh: $O^p(N)$ is the smallest normal subgroup of $p$-power index. I explain a few of its properties and put it in context in my answer. –  Jack Schmidt Oct 19 '12 at 19:23
    
@JackSchmidt: Thanks Jack, noting me that. You did below as usual the best(+1). –  Babak S. Oct 20 '12 at 6:24

1 Answer 1

Yes, there can be non-identity elements of the center of $G$.

Let $G=N$ be cyclic of order $p$. Then $O^p(N) = \langle n \in N : \gcd(p,|n|)=1 \rangle = 1$ and $G/O^p(N)$ is simple of order $p$.

For an example with $G \neq N$, take $G$ perfect of order 120, $N=Z(G)$ of order 2, and $p=3$, then $O^p(N) = N$ and $G/N$ is simple of order 60.


How can $G/O^p(N)$ be simple if $O^p(N) \neq N$? There is only one way: for $N=G$ and $G/O^p(G)$ be cyclic of order $p$. Then of course this can happen: take $G=N=S_5 \times C_3$ of order 360, and $p=2$, then $O^p(N) = A_5 \times C_3$ is order 180, and the quotient is simple of order 2.

The p-residual

If $N$ is a finite group and $\mathcal{X}$ is a class of finite groups closed under isomorphism, quotients, subgroups, and finite direct products, then $O^{\mathcal{X}}(N) = \bigcap\{ M \unlhd N : N/M \in \mathcal{X} \}$ is the unique normal subgroup of $N$ such that $N/M \in \mathcal{X}$ if and only if $M \geq O^{\mathcal{X}}(N)$. If $\mathcal{X}$ is the class of abelian groups, then $O^{\mathcal{X}}(N) = [N,N]$ is the derived subgroup. If $\mathcal{X}$ is the class of $p$-groups, then $O^{\mathcal{X}}(N) = O^p(N)$ is the subgroup in the question.

One can also view $O^p(N)$ as the subgroup generated by the Sylow $q$-subgroups for $q \neq p$, as Cauchy's theorem describes the only obstruction to being a $p$-group is having non-identity elements of order a power of a distinct prime $q$.

Obviously $O^p(N)$ is a characteristic subgroup of $N$ and contains all elements of order coprime to $p$. In particular, it can be a very large subgroup, so quotienting out by it can lose all sorts of information, including the center. Requiring $G/O^p(N)$ to be simple merely requires $N$ and $O^p(N)$ to be “large”, so it makes it very easy for them to contain central elements (which are then lost during the quotient).

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Thank you so much. If $G$ is insoluble group and $O^p(N)\neq N$, then whether there is any element in the center of $G$? –  Ashki Oct 19 '12 at 18:15
    
@Ashki, I editted the question to give an example. –  Jack Schmidt Oct 19 '12 at 19:04

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