Ignoring components of tiny order in the straightforward team of order $60$

The straightforward team of order $60$ can be created by the permutations $(1,2)(3,4)$ and also $(1,3,5)$, yet all you require to do is make even the first one and also it comes to be the identification. Can not we locate a variation of the straightforward team where the components of tiny order can be overlooked?

For a team $H$, specify $Ω_n(H)$ to be the subgroup created by components of order much less than $n$. As an example, if $n=3$ and also $H=\operatorname{SL}(2,5)$ is the excellent team of order $120$, after that $Ω_n(H)$ has order $2$, and also $H/Ω_n(H)$ is the straightforward team of order $60$. If $n=4$ and also $H=\operatorname{SL}(2,5)⋅3^4$ is the excellent team of order $(60)⋅(162)$ whose $3$-core is not enhanced, after that $Ω_n(H)$ has order $162$ and also $H/Ω_n(H)$ is once more the straightforward team of order $60$.

My first inquiry is if there are smaller sized instances for $n=4$, given that the dive $1$, $2$, $162$ appears a little bit radical for $n=2, 3, 4$.

Exists a team $H$ of order much less than $(60)⋅(162)$ such that $H/Ω_4(H)$ is the straightforward team of order $60$?

Possibly, for each and every favorable integer $n$, there is a limited team $H$ such that $H/Ω_n(H)$ is the straightforward team of order $60$. I want whether such $H$ can be picked to be "tiny" in some way.

Exists a series of limited teams $H_n$ and also a constant $C$ such that $H_n/Ω_n(H_n)$ is the straightforward team of order $60$ and also such that $|H_n| ≤ C⋅n$?

I would certainly additionally be great with some referrals to where such a trouble is reviewed. It would certainly behave if there was some type of analogue to the Schur multiplier defining the biggest non-silly bit, and also a clear definition of what a foolish bit is (I assume it is way too much to request for a non-silly bit to be had in the Frattini subgroup, and also I assume it could be unreasonable to request for the maximum among marginal bits ).


In instance it aids, below are some lowered instances that I recognize can be taken care of :

A less complex instance : if as opposed to the straightforward team of order $60$, we focus on the straightforward team of order $2$, after that we can pick $H_n$ to be the cyclic team of order $2^{1+\operatorname{lg}(n−1)}$ when $n≥2$, and also the order of $H_n$ is bounded over and also listed below by multiples of $n$. We can create a lot bigger $H_n$ for $n≥3$ by taking the straight item of our tiny $H_n$ with a primary abelian $2$-group of huge order, yet after that $Ω_n(H_n×2^n) = Ω_n(H_n)×2^n$ has actually simply come to be foolish given that the whole primary abelian $2$-group component, $2^n$, is unconnected and also makes use of a great deal of added generators, that is, it is not had within the Frattini subgroup.

A modest instance : if as opposed to the straightforward team of order $60$, we take the non-abelian team of order $6$, after that I can locate an all-natural selection of $H_n$ with $|H_n| ≤ C⋅n$, yet I am not exactly sure if there are various other practical selections. My selection of $H_n$ has $Φ(H_n)=1$, which recommends to me that Frattini expansions might not be the appropriate suggestion.

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2019-05-07 15:05:21
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