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I'm trying to understand a proof in "Groups with a maximal irredundant 6-cover" by A. Abdollahi et al, page 3231, Lemma 4.1 itens (2) and (3).

Here's my problem:

Lemma 4.1

(2) The following are $\mathfrak{G}_6$-groups:

(d) $G \cong (C_5 \times C_5) \rtimes C_4$ with $Z(G) = 1$. In this case, the intersection of a maximal irredundant cover which is core-free has size 2.

(3) A subdirect product G of three symmetric groups $Sym_3$ is a $\mathfrak{G}_6$-groups if and only if G is isomorphic to one the following groups: $(C_3 \times C_3) \rtimes C_2$, $S_3 \times S_3$, $(C_3)^3 \rtimes C_2$, where the size of the intersection of an arbitrary maximal irredundant 6-cover with core-free intersection is 1, 1, and 2, respectively.


$(*)$ G is $\mathfrak{G}_6$-groups mean $G = M_1 \cup ...\cup M_6$ is a maximal irredundant 6-cover with core-free intersection.

$(**)$ In outline proof is used the following function written with GAP (2002) program

f:= function(G) local S, M, n, C, i, T, Q, R; n:=Size(G); M:=MaximalSubgroups(G); C:=Combinations(M,6); S:=[]; for i in [1..Size(C)] do if Size(Union(C[i]))=n then Add(S, C[i]); fi; od; T := []; for i in [1..Size(S)] do if Size(Core(G, Intersection(S[i])))=1 then Add(T,S[i]); fi; od; R:=[]; for i in [1..Size(T)] do Q:=Combinations(T[i],5); if (n in List(Q, i->Size(Union(i))))=false then Add(R,T[i]); fi; od; return R; end;

The input of the function is a group G and the outputs are all the irredundant 6-covers with core-free intersection for G, and if there is no such a cover for G, the output is the empty list.

How to apply to items (2) and (3) in the function f written in the GAP to give a demonstration?

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You did not ask any question. – Mariano Suárez-Alvarez Jan 12 '11 at 21:18
Dear Moreira, it is not necessary to shout (i.e. use ALL CAPS). – Akhil Mathew Jan 12 '11 at 21:56
First, you need to know what kind of groups the function assumes (GAP has several types of groups, and not all commands know how to deal with all types of groups). Then you find a way of constructing the group (generators and relations, permutations, and so on); you can find information on doing this in Chapter 5 of the GAP tutorial; – Arturo Magidin Jan 13 '11 at 3:55
One way I often construct groups in GAP when they are given by expressions as these is: Start with the basic ones (cyclic, symmetric and so on) since these can be easily obtained. Then construct the desired direct or semidirect products. Sometimes, it might not be easy to figure out which emidirect product to choose. In those cases, one can usually (at least with groups of this size) just construct them all and check which ones satisfy the desired conditions. – Tobias Kildetoft Jan 13 '11 at 10:37

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