Abstract

Abstract It was proved by Mader that, for every integer l, every k-connected graph of sufficiently large order contains a vertex set X of order precisely l such that G − X is ( k − 2 ) -connected. This is no longer true if we require X to be connected, even for l = 3 . Motivated by this fact, we are trying to find an “obstruction” for k-connected graphs without such a connected subgraph. It turns out that the obstruction is an essentially 3-connected subgraph W such that G − W is still highly connected. More precisely, our main result says the following. For k ⩾ 7 and every k-connected graph G, either there exists a connected subgraph W of order 4 in G such that G − W is ( k − 2 ) -connected, or else G contains an “essentially” 3-connected subgraph W, i.e., a subdivision of a 3-connected graph, such that G − W is still highly connected, actually, ( k − 6 ) -connected. This result can be compared to Maderʼs result which says that every k-connected graph G of sufficiently large order ( k ⩾ 4 ) has a connected subgraph H of order exactly four such that G − H is ( k − 3 )-connected.

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