Vertex connectivity and edge connectivity of this graph Consider the following graph:   



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*The edge connectivity should be $2$. We can think this in two way:  


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*If I cut edge $(1,2)$ and $(2,4)$, the node $2$ is disconnected from the whole graph.   

*Also, edge connectivity can be thought as the network flow problem, the maximum number of edge-disjoint paths from node $1$ to node $2$ is $2$. 


*According to the property that vertex connectivity $\leq$ edge connectivity, 
the answer for vertex connectivity should be less or equal $2$.   
However, I delete node $4$ and node $6$ and all edges connecting both; the graph is still connected. Both nodes are the nodes with maximal degree.   
Where am I wrong? 
Note: the degree of node $5$ and node $7$ are $4$
 A: Vertex connectivity describes the minimum number of vertices you can remove to disconnect the graph. Not every pair needs to disconnect it in this case.
A: Let $G$ be a graph and let $\kappa(G)$ be the size of any minimum vertex separating set of $G$, $\lambda(G)$ be the minimum size of any edge separating set of $G$, and let $\delta(G)$ be the minimum degree of $G$. It is well known that $$\kappa(G)\le\lambda(G)\le\delta(G).$$
Since $\delta(G)=2$ and $G$ is connected then $\kappa(G)=1\text{ or }2$. Can you show that $\kappa(G)\ne 1$?
When you are drawing graphs, edges should never meet unless they are crossing or share a vertex in common and they should only meet in those locations. The way you have drawn your graph seems to indicate that $\deg(v_5)=3$ when it is really true that $\deg(v_5)=4$. Similarly for $v_7$.
A: If you remove vertices 1,9 and all the edges that falls on those vertices, then the vertex 11 tends to separate from the graph and hence result into disconnected graph.
So this gives edge connectivity = 2 and vertex connectivity = 2 as well.
Hence vertex connectivity <= edge connectivity. 
