Hello Math StackExchange!
A $k-$Tutte-separation of a graph $G$ is the pair $(H,K)$ of edge disjoint non-null graphs $H$ and $K$ such that $G = H \cup K$ (observe this means there are no edge crossings bw $H$ and $K$) such that $E(H) \geq k$ and $E(K) \geq k$, and $|V(H) \cap V(K)| = k$. A graph $G$ is said to be $k-$Tutte-connected if there exist no $m-$Tutte-separations such that $m < k$.
I am working on this problem from “A course in Combinatorics” by JH Van Lint, RM Wilson.
Problem $\bf 32D$ (van Lint and Wilson) Prove that if $|V(G)|\ge k+1$, then $G$ is $k$-Tutte connected if and only if $G$ is $k$-vertex connected and has no polygons with fewer than $k$ edges.
Here, $k-$vertex connectivity is defined as “the standard one” on Wikipedia. I am very stuck on proving that if $G$ contains a $m-$gon ($C_m$) such that $m < k$, then $G$ is not $k$-Tutte-connected. I have tried everything to the extent I am now exhausted.
PS : Yes, Tutte is infamous for his unconventional definitions.
If we can find an $m$-cycle, then we usually get an $m$-Tutte-separation by taking $H$ to be the cycle subgraph and $K = G-E(H)$. This satisfies $|E(H)|=m$ and $|V(H) \cap V(K)| = m$, so if there are at least $m$ edges in $K$, then all the conditons hold.
It's possible that $|E(K)|<m$. In this case, suppose there is a vertex $v$ with degree $d$ such that $d \le \frac12 |E(G)|$; note that $m > \frac12 |E(G)|$, so $d \le m$. Let $H$ be the subgraph with $V(H) = \{v\} \cup N(v)$ keeping only the edges incident to $v$, and let $K = G-v$. This is a $d$-Tutte-separation: $|E(H)|=d$, $|E(K)| \ge d$ because there are at least $2d$ edges total, and $|V(H) \cap V(K)| = |N(v)| = d$.
Finally, if all vertices have minimum degree more than $\frac12|E(G)|$, there are at most $3$ vertices; the graph is either $K_2$ or $K_3$, and there's nothing to worry about here.