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AdrianZ
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Let G be a non-cyclic group of order pn where p is a prime number. Prove that G has at least p+3 subgroups.
Could anyone offer a solution to this problem?
Could anyone offer a solution to this problem?
morphism said:Here's a quick proof. Let [itex]\Phi(G)[/itex] be the Frattini subgroup of G (= intersection of maximal subgps of G). Then [itex]G/\Phi(G) = (\mathbb Z/p\mathbb Z)^k[/itex] for some [itex]k\geq 1[/itex] and moreover k=1 iff G is cyclic. Since G isn't cyclic, we see that [itex]G/\Phi(G)[/itex] is an elementary abelian p-group of rank at least 2, hence has at least (p^2-1)/(p-1)=p+1 subgroups of index p. Consequently G has at least p+1 subgroups of index p. Now throw in the trivial subgroup and G itself to see that G has at least p+3 subgroups.
micromass said:Sigh... I was hoping that the OP would discover the answer by our hints...
The different proofs are all nice though.
I'd say graduate-level, if at all.Deveno said:by gosh, that's slick...let me ask you this, at what level do you think students get their first exposure to the Frattini subgroup?
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