What's an easy way of proving a subgroup is normal? I think in most situations(for example, in $S_n$ or $D_n$), proving by definition is too complicated because you have to calculate $gng^{-1}$ for every $n$ in $N$ and $g$ in $G$. To prove that all the left cosets are also right cosets is also too complicated because you have to find all those cosets. I wonder if there's a way to do this without having to calculate everything by hand.
 A: If you know a subgroup of a particular order is the ONLY subgroup of that order, then you know it's normal. I know that's a unique case, but just another tool to keep in mind.
A: There are a number of ways in which the work can be shortened.


*

*If you can come up with a homomorphism whose kernel is precisely $N$, then this guarantees that $N$ is normal. This is often the case.

*It suffices to check a generating set for $N$. That is, if $N=\langle X\rangle$, then $N$ is normal in $G$ if and only if $gxg^{-1}\in N$ for every $x\in X$. For instance, this makes proving that the subgroup generated by all $m$ powers is normal easy.

*It suffices to check a generating set for $G$ and its inverses. That is, if $G=\langle Y\rangle$, and $yNy^{-1}\subseteq N$ and $y^{-1}Ny\subseteq N$ for all $y\in Y$, then $N$ is normal.
A: That would depend on the problem. I believe the following properties are most useful.
A subgroup $N$ of $G$ is normal iff one of the following is true:


*

*For every $g\in G$ and $n\in N$, $gng^{-1}\in N$.

*For every $g\in G$, $gNg^{-1}\subseteq N$.

*For every $g\in G$, $gNg^{-1}=N$.

*Every left coset of $N$ is a right coset of $N$.

*The product of two right cosets of $N$ is again a right coset of $N$.

A: If your subgroup has index 2, then it is always normal (because whether you consider left or right cosets, there are only these 2: the subgroup itself, and the rest of the elements).
Another way (maybe the best way) is to show that the subgroup is the kernel of a homomorphism having the group as its domain.
