Constructing Steiner System $S(5,8,24)$ and Mathieu Group $M_{24}$ Let $a_0,...,a_{2^{24}-1}\in \{0,1\}^{24}$ be a sequence such that $a_k$ is the binary representation of $k$. i.e.
$$a_0=(0,...,0,0,0,0)$$
$$a_1=(0,...,0,0,0,1)$$
$$a_2=(0,...,0,0,1,0)$$
$$a_3=(0,...,0,0,1,1)$$
$$.$$
$$.$$
Let $b_0,...,b_{2^{12}-1}$ be a subsequence of $a_0,...,a_{2^{24}-1}$ such that
$$b_0=a_0$$
and for $n\geq 1$, 
$$b_n=a_k$$
where $k$ is the least number such that $a_k$ differs in at least 8 components 
from each of $b_0,...,b_{n-1}$.
why is the sequence $b_k$ well-defined?
Let $L$ be the set of all $b_k$ that have exactly 8 nonzero componenets. Let
$$X=\{1,2,...,24\}$$
Clearly each $b\in L$ is in one-one correspondence with an 8-element subset $S_b\subseteq X$. Let
$$\mathcal{B}=\{S_b|b\in L\}$$
I'm trying to show that $(X,\mathcal{B})$ is an $S(5,8,24)$ steiner system. 


*

*$|X|=24$.

*$(\forall B\in\mathcal{B})(|B|=8)$.

*$(\forall A\in X^{\{5\}})(\exists! B\in\mathcal{B})(A\subseteq B)$.


in 3 why does $B$ exist? (it's easy to show that it's unique if it exists).
Let
$$M_{24}=Aut(X,\mathcal{B})$$
why does $M_{24}$ act 5-transitively on $X$?
 A: I'll answer all your questions except the 5-transitivity of $M_{24}$.  That would require constructing symmetries of $(X, \mathcal{B})$, which is very hard to do using this construction (there are no "obvious" symmetries in this construction to get us started).
Let $V$ be the set of subsets of $X$, and let $C$ be the subset of $V$ corresponding to the $b$'s.  Then $V$ is a 24-dimensional vector space over $\mathbb{F}_2$ (the sum of two subsets, $A+B$, is the set of elements of $X$ that lie in exactly one of $A$ or $B$).
Claim 1: $C$ is a subspace of $V$.  
This claim is the most difficult part of the proof, and we'll come back to it at the end. 
Note that $|V| = 2^{24}$ and $|C| = 2^{12}$, so that the quotient group $V/C$ also contains $2^{12}$ elements.
Claim 2: $C$ does not contain any nonempty subsets of $X$ of size less than 8.
Proof: By construction, the empty set lies in $C$ (it corresponds $b_0 = 0$) and any other element of $C$ differs from the empty set in at least 8 positions.
Claim 3: Suppose $S$ and $T$ are subsets of $X$ having size $\le 4$.  If $S+T \in C$, then $|S|=|T|=4$, and $S$ and $T$ are disjoint.
(Proof: This follows easily from Claim 2.)
Choose $x \in X$.   Let $\mathcal{R}$ be the set of all subsets of $X$ of size $\le 3$ together with the set of all subsets of $X$ of size 4 containing $x$.  (Note that $\mathcal{R}$ depends on $x$, but we will suppress the dependence on $x$ in the notation.)
Claim 4: The set $\mathcal{R}$ is a complete set of coset representatives for $C$ in $V$.  In other words, for every subset $S$ of $X$, there is a unique $T \in \mathcal{R}$ such that $S+T \in C$.
Proof: By Claim 3, no two elements of $\mathcal{R}$ lie in the same coset of $C$ in $V$.  But the number of elements of $\mathcal{R}$ is $$\binom{24}{0} + \binom{24}{1} + \binom{24}{2} + \binom{24}{3} + \binom{23}{3}$$ which, miraculously, equals $2^{12}$, the number of cosets of $C$ in $V$.  Therefore, every coset contains a unique element of $\mathcal{R}$.
Claim 5: The set of elements of size 8 in $C$ form a $(5,8,24)$ Steiner system.
Let $S$ be a subset of $X$ of size 5.  Choose $x \in S$, and construct $\mathcal{R}$ as above.  By claim 4, there exists a unique $T \in \mathcal{R}$ such that $S+T \in C$.  This $T$ must have size 3, because size 0,1, or 2 would clearly be too small (by claim 2), and size 4 does not work because such a set must contain $x$, which is also in $S$, so again the sum would have size $\le 7$.  Also, $S$ and $T$ are disjoint, or else $S+T$ would again be too small.  Therefore $S+T$ is a set of size 8 in $C$ containing $S$.  The uniqueness of $T$ implies that this 8-element set in $C$ is unique.
Reference for all of the above.
Okay, now back to claim 1.  This will only be a sketch.  For details, look up the theory of impartial combinatorial games. 
We consider the following two-player game.  The starting position is an integer $n$, with $0 \le n \le 2^{24}-1$.  Players alternate turns.  A legal move is to replace an integer $n$ with a smaller nonnegative integer $m$ such that $m$ and $n$ differ by no more than 7 bits.  The player that moves to 0 wins.  (Reference: This is the game "Turning Turtles" in Winning Ways for your Mathematical Plays, by Berlekamp, Conway, and Guy, Volume 3 (2nd ed.), Chapter 14.  In the 1st edition, it's in Volume 2.)
I claim that the values $b_i$ which make up $C$ are precisely the starting integers for which the second player can win.  Proof: by induction.  If the starting position is 0, then the second player has won by definition.  If the starting position $n$ is in between 2 consecutive $b_i$ and $b_{i+1}$, then $n$ and $b_i$ differ by at most 7 bits, for if they differed by 8 or more, this would contradict the construction of $b_{i+1}$.  Therefore, it is legal for the first player to move from $n$ to $b_i$, and now the first player becomes the second player in the new game $b_i$ which (by the inductive hypothesis) is a second player win.  Therefore the first player wins $n$.  Finally, if $n = b_i$, then the first player cannot move to any smaller $b_j$ because they all differ by 8 or more bits by construction.  Therefore any move by the first player leads to a position between 2 consecutive $b$'s, which the first player (who was the second player in the original game) can win by the inductive hypothesis.
In the language of impartial combinatorial games, the numbers $b_i$ are precisely the $P$-positions of this game.
Now the theory of impartial combinatorial games tells us that there exist integers $x_0, x_1, \ldots, x_{23}$ such that if $n = c_i 2^i$ is the binary expansion of $n$, then the nim sum $\oplus c_i x_i = 0$ iff $n$ is a second player win.  The nim sum is exactly the same operation as the symmetric difference of subsets in our original space $V$.  So the elements of $C$ are precisely those for which $\oplus  c_i x_i = 0$, and this latter description of $C$ is clearly linear.  Therefore, Claim 1 holds.
