Abstract Algebra - Permutations I'm asked to show that $(1,2,3) \in S_3$ generates a subgroup which is normal.
I know that I could show it explicitly but that would be tedious. I think it may have to do with the fact that $(1,2,3)$ generates all even permutations but I'm sure there's something I'm missing. Any help would be appreciated.

Actually, showing it directly would be relatively untedious as far as getting your hands dirty with group theory is concerned. There are only six elements in $S_3$, and $\langle(123)\rangle$ has three elements so it only has two cosets (where did I get "two" from?), so there isn't much work involved. I urge you to do the problem first by going this route. This kind of practice is necessary.
For an abstract, and perhaps elegant, approach, you can argue that $\langle(123)\rangle$ is the only subgroup with some property (and explain why it is the only one), a property which is invariant under conjugation. This will tell you the subgroup is conjugation-invariant, i.e. normal (quiz: how does it tell us this?). Can you figure out the property? It's very basic: I mentioned it in the first paragraph.
It's possible that the meat of this argument (how being the unique subgroup with a conjugation-invariant property implies the subgroup is conjugation-invariant) is what you're struggling with. But you have the right idea since the argument I gave above generalizes to showing $A_n\triangleleft S_n$ using the same argument with even permutations (although one could also show it's index two).
