By Kuratowski's theorem, every nonplanar graph contains a (topological) minor of $K\_5$ or $K\_{3,3}$.  

But I observed that every time I construct a $4$-connected nonplanar graph, it always contains not only a $K\_{3,3}$-minor but also a $K\_5$-minor.  

Moreover, although I tried many times, I cannot construct a $4$-connected nonplanar graph containing only $K\_{3,3}$-minors!  

So I want to know whether the following statement is true:



> 
> Every $4$-connected nonplanar graph contains a $K\_5$-minor.
> 
> 
> 


Unfortunately, I could not find any references about this topic.  

If the statement is true, can you give me a proof?  

Or if it's not, can you show me a counterexample?  

Thanks a lot.



Does every $4$-connected nonplanar graph contain a $K_5$-minor?
Yes, this is true and follows from [Wagner's theorem](https://en.wikipedia.org/wiki/Wagner%27s_theorem). Wagner's theorem asserts that every graph with no $K\_5$ minor can be built from $0$-, $1$-, $2$-, and $3$-sums from planar graphs and a fixed $8$ vertex non-planar graph called the [Wagner graph](https://en.wikipedia.org/wiki/Wagner_graph). Since the Wagner graph is not $4$-connected, this implies that every $4$-connected graph with no $K\_5$ minor is planar.


Alternatively, there is a short proof provided you are happy to assume Kuratowski's theorem. The proof idea is to start with a model of $K\_{3,3}$ and then use $4$-connectivity to 'augment' the model of $K\_{3,3}$ to a model of $K\_5$. See the paper [A Quick Proof of Wagner's Equivalence Theorem](https://londmathsoc.onlinelibrary.wiley.com/doi/10.1112/jlms/s2-3.4.661) by Young.


