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Let (N_{0}) be a maximal subgroup of (N).
Let V be a maximal subgroup of G such that (Lle V).
Therefore, X is not a maximal subgroup of H. Let M be a maximal subgroup H such that M contains X.
Now, let M be a maximal subgroup of K such that |M|= q γ r β where 0≤β≤α and 0≤γ≤1.
If not, then I G < G. Let U containing I G be a maximal subgroup of G. Then 1 ≤ | G : U | ∣ | G : I G | = 4.
Suppose that ((Hbigcap K K_{p^{prime }}) is a proper subgroup of (K) and let (K_{0}) be a maximal subgroup of (K) containing ((Hbigcap K K_{p^{prime }}).
Similar(51)
Hence (Mbigcap K) is a maximal subgroup of (K).
Moreover (Mbigcap H) is a maximal subgroup of (H).
Hence there is a maximal subgroup (Lne N) of (K) such that (K=NL).
Since (M_{1}^{h}) is a maximal subgroup of (H), it follows that (M_{1}^{h}) permutes with (M_{2}).
In Section 5, we show that for a word wi of weight i, where i∈{100,112,164,176} the stabilizer (McL wi is a maximal subgroup of McL.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com