A Hamiltonian graph $G$ of order n is $k-$ordered Hamiltonian for an integer $k$ with $1≤k≤n$ if for every ordered set $S=\{v_1,v_2,…,v_k\}$ of $k$ vertices of $G$, there is a Hamiltonian cycle of $G$ encounter these $k$ vertices of $S$ in the order listed.
a) Let $G$ be a graph of order $n≥3$ such that $deg(v)≥\frac{n}{2}$ for every vertex $v$ of $G$. Show that $G$ is $3-$ordered Hamiltonian.
b) Let $G$ be a graph of order $n≥4$ such that $deg(v)≥\frac{n}{2}$ for every vertex $v$ of $G$. Show that $G$ need not be $4-$ordered Hamiltonian.
c) Show that if $G$ is 4-ordered Hamiltonian graph, then $G$ is 3-connected.
For both a) by Dirac's theorem, I know that $G$ is Hamiltonian, meaning $G$ has a cycle that go through every vertex of $G$. So Let $S=\{a,b,c\}$. We can find a cycle that goes through $S$ by following orders : $abc, acb, bca,bac, cab,cba$. So $G$ is 3-ordered Hamiltonian
for c), I'm still stuck
Proof of part c)
Note that for c) we need the additional requirement that $|G|>3$: $K_1$ certainly is 4-ordered Hamiltonian according to the given definition, but it is not 3-connected.
Suppose $G$ is not 3-connected. Then it has a vertex cut of size 2, say $\{x,y\}$. Suppose this vertex cut separates vertices $a$ and $b$. Because $G$ is 4-ordered Hamilton, there is a cycle in $G$ that contains the vertices $a,x,y,b$ in this sequence. Now by walking from $a$ to $b$ along this cycle, using the path that avoids $x$ and $y$ we have exhibited an $a,b$-path in $G-\{x,y\}$. Contradiction.