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Full question: Let $G$ be a group that acts $t$-transitively on a set $X$. Let $S$ be a $k$-subset of $X$. Show that the orbit $S^G$ of $S$ under the action of $G$ forms a $t-(n,k,\lambda)$-design with $\lambda=\frac{|G|\binom{k}{t}}{|Stab_S|\binom{n}{t}}$.

My work/issues: For any $t$-design, I know that $\lambda=b\frac{\binom{k}{t}}{\binom{n}{t}}$, where $b$ is the number of blocks. So I know that $b$ somehow has to be equal to $\frac{|G|}{|Stab_S|}$. If the orbit $S^G$ forms a t-design, then the number of elements of $S^G$ should be the number of blocks. So I'm interested in $|S^G|$.

Now, the Orbit Counting Lemma says that $|S^G|=\frac{1}{|G|}\sum_{x\in S}|Stab_x|$, but I don't see how this can lead be to $\frac{|G|}{|Stab_S|}$. I think the key to the problem is making use of how $G$ acts $t$-transitively on $X$, but I don't see how to make use of it!

Also, if it is true that $|S^G|$ corresponds to $b$, then the size of each block must be $k$, which would correspond to the size of each orbit of $S$ (i.e. $|orb_G(i)|$, $1\leq i \leq k$). This I also struggle to see.

Thanks a lot for any assistance!

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I presume you want $k>t$. For each $t$-set $A$, if $A$ is contained in exactly $\lambda$ elements of $S^G$, namely $B_1,\ldots,B_\lambda$ then $A^g$ (for $g\in G$) is also contained in exactly $\lambda$ elements of $S^G$, namely $B_1^g,\ldots,B_\lambda^g$. As $G$ is transitive on $t$-sets, then $S^G$ is a $t$-design for some $\lambda$.

What is $\lambda$? The number of blocks is $b=\lambda{n\choose t}{k\choose t}^{-1}$. But $b=|G:\text{Stab}_G(S)|$ (just the orbit-stabiliser theorem). This gives the formula you seek.

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