this is a strange problem! the reason is that i think if is characteristically simple. that's because is a copy of for anywhy?thus

but is a non-trivial characterstic subgroup of thus because is characteristically simple.

do you know this fact that the Frattini subgroup, is the set of all the non-generators of G? if you know this, then your problem is not that hard. first note that if G has no maximal subgroup

The second question is to show that if N is normal in G then where is the intersection of all maximal groups in A.

(this might happen if G is not finite), then by definition and there's nothing to prove. otherwise, if for any maximal subgroup M of G, then we're done. so suppose there

exists a maximal subgroup M of G such that now since is characterstic in N and N is normal in G, is normal in G. thus is a subgroup of G properly containing M.

thus by maximality of M. thereforewhy?that means and hence because the elements of are non-generators.

so we've proved that which gives us contradiction!