Prove that $G_a$ has a hamilton circuit So I need some help with this problem.
I thought I used the right theorem but dont think so.
Here is my try:
We have graph G, $a\in$ N and define $G_a$:


*

*$V(G_a)=\{(x,p):v\in V(G), 1\leq p\leq a\}$

*$\{u,v\}\in E(G), 1\leq p\leq a \iff \{(u,p),(v,p)\}\in E(G_a)$ and $v\in V(G), 1\leq p\leq a-1 \iff \{(v,p),(v,p+1)\} \in E(G_a)$.
Prove that $G_a$ has a hamilton circuit if $3\leq |G|$ and G has such a circuit 
My attempt:
I know by a theorem that if G is a graph and $|G|=n\leq 3$ and $\delta(G)\leq n/2$. So first of all $3\leq |G_a|$. I know want to prove that $\delta(G_a)\geq n/2$. How can I proceed?
 A: Revised to get a Hamiltonian circuit.
You can describe a Hamiltonian path in $G_a$ directly. Think of $G_a$ as a vertical stack of $a$ copies of $G$, with vertical edges between corresponding vertices in adjacent copies. Let $P$ be a Hamiltonian cycle $v_0v_1\ldots v_nv_0$in $G$.
Start at vertex $\langle v_0,1\rangle$ in the top copy and trace the path in that copy to $\langle v_n,1\rangle$. Drop down to the $\langle v_n,2\rangle$ in the second copy and trace it in reverse, but only as far as $\langle v_1,2\rangle$. Drop down to $\langle v_1,3\rangle$ in the third copy and trace it forward to $\langle v_n,3\rangle$. Continue in this fashion. 
If $a$ is odd, you’ll trace the path forward in level $a$, ending at $\langle v_n,a\rangle$, and you can complete the circuit by going from there to $\langle v_0,a\rangle$ and then straight up through the vertices $\langle v_0,k\rangle$ until you finish back at $\langle v_0,1\rangle$.
If $a$ is even, you’ll trace the path in reverse in level $a$, ending at $\langle v_1,a\rangle$, and you can complete the circuit by going to $\langle v_0,a\rangle$ and then straight up through the vertices $\langle v_0,k\rangle$ until you finish back at $\langle v_0,1\rangle$.
