A question about $n$-cubes and boolean lattices (graph theory). From Graph Theory by Adrian Bondy and U.S.R. Murty:

1.1.7 The $n$-cube $Q_n$ ($n \ge 1$) is the graph whose vertex set is the
  set of all $n$-tuples of $0$s and $1$s, where two $n$-tuples are adjacent
  if they differ in precisely one coordinate.
a) Draw $Q_1$, $Q_2$, $Q_3$, and $Q_4$.
b) Determine $v(Q_n)$ and $e(Q_n)$.
c) Show that $Q_n$ is bipartite for all $n \ge 1$.
1.1.8 The boolean lattice $BL_n$ ($n \ge 1$) is the graph whose vertex set
  is the set of all subsets of $\{ 1, 2, \ldots , n \}$, where two
  subsets $X$ and $Y$ are adjacent if their symmetric difference has
  precisely one element.
a) Draw $BL_1$, $BL_2$, $BL_3$, and $BL_4$.
b) Determine $v(BL_n)$ and $e(BL_n)$.
c) Show that $BL_n$ is bipartite for all $n \ge 1$.

When trying to give a solution I found great resemblance between the two (in fact, I found that they're identical, though I'm unsure about this). So my questions are:


*

*How do I solve part c)?

*Are they ($Q_n$ and $BL_n$) really identical (isomorphic) and why?


I think boolean lattice is related to abstract algebra but I don't know about that. So please state your answers using graph-theoretic notations and concepts. Thanks in advance.
 A: You can define a bijection between $n$-tuples on $\{0,1\}$ and subsets of an $n$-element set. The basic idea is that the tuple is the characteristic function of the subset, or in other words, we include an element in the subset if the corresponding position in the tuple is non-zero.
Define $\phi: \{0,1\}^n \to \mathcal{P}(\{1,2,\ldots,n\})$ as
$$\phi\Big((x_1,x_2,\ldots,x_n)\Big) = \Big\{ i \in \{1, 2, \ldots, n\} \ \Big|\ x_i \neq 0\Big\}.$$
Indeed, $\phi$ can be made into a graph isomorphism.
To prove that the first graph is bipartite, color the graph according to
$x_1 \oplus x_2 \oplus \ldots \oplus x_n$, i.e., the number of non-zero entries modulo $2$. 
To prove that the second graph is bipartite, prove that it has no triangle, that is, for three sets $A$, $B$, $C$ such that $|A \oplus B| = 1 = |B \oplus C|$ we have $|A \oplus C| \neq 1$. In fact this is really the same proof as the previous one, only obscured: consider the number of elements of $A$, $B$ and $C$, and observe that at least two have to be odd or at least two have to be even.
I hope this helps $\ddot\smile$
