Lemma 74 ([Taking cores without losing a simple image]). Let R_0≤D , let C R_0 , and suppose that R_0/C is nonabelian simple. Let L_1,…,L_s≤D be normalized by R_0 . Suppose that, for every t , there is a subgroup D_t R_0 such that …

Let π:R_0→R_0/C be the quotient map. Start with B=R_0 . At every stage B will be normal in R_0 and satisfy π(B)=R_0/C . For a fixed t , replace B by F_t(B)=⋂_ ℓ∈L_t (B∩D_t)^ℓ, K^ℓ=ℓ^ -1 Kℓ. The subgroup B∩D_t still maps onto R_0/C by Lemma 70 ([Intersections retaining a perfect image]). Put K=B∩D_t . The subgroup K is normal in D_t , because B R_0 and D_t≤R_0 . Since L_t normalizes D_t , every conjugate K^ℓ lies in D_t and is normal in D_t . For d∈D_t and ℓ∈L_t , the assumption [D_t,L_t]≤D_t∩C gives π(d^ℓ)=π(d) . Consequently π(K^ℓ)=π(K)=R_0/C for every ℓ . This applies the quotient map only to elements of D_t≤R_0 ; it does not require C to be normalized by L_t . Applying Lemma 70 ([Intersections retaining a perfect image]) inside D_t shows that F_t(B) has full image. The new subgroup is normal in R_0 . Indeed, R_0 normalizes B , D_t , and L_t . For r∈R_0 one therefore has (K^ℓ)^r=(K^r)^ ℓ^r =K^ ℓ^r , and ℓ^r∈L_t . Thus R_0 permutes precisely the conjugates intersected in. The operation is descending, since the identity conjugate occurs. Apply F_1,…,F_s in repeated rounds. Finiteness forces stabilization, and at a stable round every intermediate inclusion is equality. The resulting subgroup B_* is therefore normalized by each L_t as well as by R_0 . Thus B_* W and B_*≤R_0 , while π(B_*)=R_0/C . Since B_*_W(R_0) , the asserted core also has full image.