I was reading up on this: Find the number of normal subgroups of $A_4$.
If $H$ has a $3$-cycle, say $(123)$, then $H$ has its inverse $(132)$ thefore it also has $(124) = (324)(132)(324)^{-1}$, Thus $H \supseteq \langle (123),(124) \rangle = A_4$ (why?). Conclusion $H = A_4$.
How can we claim the above without loss of generality? How can we be sure that for some other choice of $3$-cycles we get the same result?
Well, I think his point is to have you understand the idea and work out the details through analogy (which is sometimes source of errors in proofs, but not in this case). A more precise exposition would be:
Let's say $H$ contains a $3$-cycle $\sigma=(abc)$. Let's call $d$ the element of $\{1,2,3,4\}$ which does not appear in $\{a,b,c\}$.
Then $\sigma^{-1}=(acb)\in H$. And $(abd)=(cbd)(acb)(cbd)^{-1}\in H$ by normality.
Thus $H\supseteq\langle (abc),(abd)\rangle=A_4$, because the group generated by those two $3$-cycles must still contain $(abc)(abd)=(ac)(bd)$, hence at least $8$ elements.
Let's say $H$ contains a $3$-cycle $(abc)$ where $d$ is the element left out. Then, $H$ has its inverse, $(acb)$, so it also has $(abd)=(cbd)(acb)(cbd)^{-1}$. Thus, $H \supseteq \langle (abc), (abd) \rangle=A_4$. We now conclude that $H=A_4$.
I'm not sure if this is how permutations work since I'm not really good with permutations, but I'm guessing this is how they assumed loss of generality.
Two $3$-cycles on $4$ elements that don't share a fixed point necessarily generate a transitive subgroup $G$ of $A_4$. Therefore, by the orbit-stabilier theorem, the order of $G$ is divisible by four. Because $|G|$ must also be divisible by $3$ (Lagrange). Therefore $12\mid |G|$, and the claim follows.
Alternatively we can also deduce that $G$ must be doubly transitive, because the presence of a $3$-cycle shows that a point stabilier is transitive. Again, $12\mid |G|$ and we are done.
