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.
 A: 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).
A: I think that the standard way to do that is simply by brute force.
You have just to check 3 conjugations of the subgroup by the 3 elements outside of it, process that is neither hard nor long.
Anyway, a not-so-tedious way to do that is to show that it is the kernel of the (surjective) group homomorphism
$$\begin{aligned}\phi: S_3&\longrightarrow C_2\\
\sigma&\longmapsto \mathrm{sgn}(\sigma)
\end{aligned}$$
Source: Take a look at this Wikipedia page.
