Show that $\kappa(Q_n) =\lambda (Q_n)=n$ all positive integer $n$ Show that $\kappa(Q_n) =\lambda (Q_n)=n$ all positive integer $n$
I want to prove this by induction. so I start with $n=1$
Base: $n=1$ then $Q_1=K_2$, which have $\kappa(Q_1) =\lambda (Q_1)=1$. So the base is good.
Inductive step:
Assume that this work for $n=k$, meaning for $Q_k = K_2 \times K_2 \times \dots \times K_2$, $k$ times, $\kappa(Q_k) =\lambda (Q_k)=k$. I need to show that it work for $n=k+1$. I know that $Q_{k+1} = K_2 \times K_2 \times \dots \times K_2$, $k+1$ times, so $Q_{k+1}$ has $2^{k+1}$ vertices and we obtain $Q_{k+1}$ by taking 2 copies of $Q_k$ and connect their vertices pairwise, so there must be at least $k+1$ bridges. In each bridge, there must be at least one cut vertices, so  there are at least $k+1$ cut vertices in $Q_{k+1}$, So  $\kappa(Q_{k+1}) =\lambda (Q_{k+1})=k+1$
On inductive step, I don't feel like I explain it clear enough. I would be appreciate if anyone help me improve it.
 A: Yes, the induction step needs a bit of work. Assuming that you know the result that $\kappa(G)\le\lambda(G)$ for all graphs $G$, you can simplify the overall argument a bit: you need only show that $\kappa(Q_n)\ge n$ and $\lambda(Q_n)\le n$ for all $n\in\Bbb Z^+$. The latter is pretty easy and doesn’t need induction; can you see how to remove $n$ edges from $Q_n$ to disconnect it?
For the other inequality, prove by induction on $n$ that if $u$ and $v$ are distinct vertices of $Q_n$, there are at least $n$ paths from $u$ to $v$ such that if $P_0$ and $P_1$ are any two of these paths, $u$ and $v$ are the only vertices that are on both $P_0$ and $P_1$. Suppose that you’ve proved that. Then removing fewer than $n$ vertices from $Q_n$ must leave a connected graph, because if $u$ and $v$ are any two of the remaining vertices, there is a path from $u$ to $v$ in $Q_n$ that does not include any of the removed vertices.
It helps to label the vertices of $K_2$ as $0$ and $1$; since $Q_n$ is the product of $n$ copies of $K_2$, its vertices correspond to $n$-tuples of $0$s and $1$s, and two vertices are adjacent in $Q_n$ if and only if they differ in exactly one coordinate. Suppose that there are at least $n$ disjoint paths between any two vertices of $Q_n$, and consider distinct vertices $u=\langle u_1,\ldots,u_{n+1}\rangle$ and $v=\langle v_1,\ldots,v_{n+1}\rangle$ in $Q_{n+1}$. Suppose that there is some $k$ such that $u_k=v_k$. Let $u'=\langle u_1,\ldots,u_{k-1},u_{k+1},\ldots,u_{n+1}\rangle$ be obtained from $u$ by deleting the $k$-th coordinate, and similarly let $v'=\langle v_1,\ldots,v_{k-1},v_{k+1},\ldots,v_{n+1}\rangle$; these are distinct vertices in $Q_n$, so there are $n$ disjoint paths between them in $Q_n$.


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*Show how to turn these $n$ paths into disjoint paths from $u$ to $v$ in $Q_{n+1}$. HINT: What should the $k$-th coordinate of each vertex on these paths be?  

*Find one more path from $u$ to $v$ in $Q_{n+1}$. HINT: the first vertex after $u$ should be $\langle u_1,\ldots,u_{k-1},?,u_{k+1},\ldots,u_{n+1}\rangle$.


There remains the possibility that $u$ and $v$ differ in every coordinate. You might as well assume for simplicity that they are $\langle 0,\ldots,0\rangle$ and $\langle 1,\ldots,1\rangle$, since you can always rearrange the labelling of the factors $K_2$ to make this true.


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*Find $n+1$ disjoint paths from $\langle\underbrace{0,\ldots,0}_{n+1}\rangle$ to $\langle\underbrace{1,\ldots,1}_{n+1}\rangle$.

