If $G$ acts transitively on a set $A$, how can $A$ contain a block $B$? I'm having trouble understanding an exercise in Dummit and Foote.
Section 4.1, Exercise 7:
Let $G$ be a transitive permutation group on the finite set $A$. A block is a nonempty subset $B$ of $A$ such that for all $\sigma \in G$ either $\sigma(B) = B$ or $\sigma(B) \cap B = \emptyset$ (here $\sigma(B)$ is the set $\{\sigma(b) \, | \, b \in B\}$).
My question is, if $G$ acts transitively on a set $A$, how can $A$ contain a block $B$? Suppose $B$ is a block. Take any element $b \in B$ and $a \in A \setminus B$. The action is transitive, so there exists $\sigma \in G$ such that $\sigma(b) = a$, so $\sigma(B) \neq B$. Now take an element $c \in B$. There then exists $\tau \in G$ such that $\tau(c) = b$, so $\tau(B) \cap B \neq \emptyset$. But then neither condition of a block is satisfied. Thus a block cannot exist if the action is transitive. 
Help?
 A: You are parsing the definition incorrectly.
The statement says that for all $\sigma$, either $\sigma(B)=B$ or $\sigma(B)\cap B=\varnothing$. Formally, this says:
$$\forall \sigma\Bigl( \bigl( \sigma(B)=B\bigr)\vee \bigl(\sigma(B)\cap B=\varnothing\bigr)\Bigr).$$
You are interpreting it as
$$ \Bigl( \forall \sigma\bigl(\sigma(B)=B\bigr)\Bigr) \vee \Bigl( \forall \sigma \bigl( \sigma(B)\cap B=\varnothing\bigr)\Bigr).$$
The two statements are not equivalent: every hand is either a left hand or a right hand; that’s not the same as every hand is a left hand or every hand is a right hand.
(The second formula implies the first, but the first does not imply the second.)
Had the second interpretation been intended, it would have been written as: “either for all $\sigma\in G$ we have $\sigma(B)=B$, or else $\sigma(B)\cap B=\varnothing$ [for all $\sigma$]” (I can see the bracketed segment fragment being omitted). Note the placement of the “either” relative to the “for all.”
These kind of subtle language signals are pretty common, and they are hard to spot at first, so don’t feel too bad that you missed it. It’s important to keep an eye out for them.
I will also note that you are assuming $B\neq A$; that is unwarranted. There are two “trivial” cases of blocks: $B=A$ is always a block (here we do have that $\forall\sigma (\sigma(B)=B)$;  but you did not allow for that possibility when taking $a\in A\setminus B$), The other trivial case are singletons: if $B=\{a\}$, then either $\sigma(a)=a$ and $\sigma(B)=B$; or $\sigma(a)\neq a$ and $\sigma(B)\cap B=\varnothing$. We are generally interested in cases in which neither is true (when those are the only kinds of blocks that can be defined in a transitive action, we say the action is “primitive”).
