TITLE: Techniques for showing that a subgroup is not normal
QUESTION [11 upvotes]: To show that a subgroup of a group is normal, I typically construct a homomorphism whose kernel is that subgroup. Are there any general principles or tests that I can use to determine that a subgroup is not normal? Or are such claims usually proved by ad hoc methods?
In particular, I would like to show that $\textbf{SL}_n(\mathbb{Z})$ is not a normal subgroup of $\textbf{SL}_n(\mathbb{R})$.
Any help is appreciated!

REPLY [3 votes]: Given any subgroup $H$ of a group $G$, you can describe its conjugate $gHg^{-1}$ conceptually as the stabilizer of the coset $gH$ with respect to the action of $G$ on $G/H$. The intersection $\bigcap_{g \in G} gHg^{-1}$ of these conjugates is then the kernel of this action. You can sometimes identify in fairly explicit terms the action of $G$ on $G/H$, and from there it's sometimes easy to compute this intersection (and verify that it's smaller than $H$) even if it's hard to describe the conjugates of $H$ explicitly.
In your case, we can say the following. In terms of the action of $SL_n(\mathbb{R})$ on $\mathbb{R}^n$, you can define $SL_n(\mathbb{Z})$ to be the subgroup stabilizing the standard lattice $\mathbb{Z}^n$ in $\mathbb{R}^n$. $SL_n(\mathbb{R})$ acts transitively on all lattices in $\mathbb{R}^n$ with stabilizer $SL_n(\mathbb{Z})$, and so the quotient $X = SL_n(\mathbb{R})/SL_n(\mathbb{Z})$ can be identified with the space of lattices in $\mathbb{R}^n$ (not up to rotation or scaling or anything). The nontrivial conjugates of $SL_n(\mathbb{Z})$ can then be identified with the stabilizer of other lattices in $\mathbb{R}^n$. 
This action is very close to being faithful: I think when $n$ is even the kernel is generated by $-I$ and when $n$ is odd the kernel is just trivial. This is just the statement that any matrix in $SL_n(\mathbb{R})$ that isn't $\pm I$ fails to stabilize at least one lattice. If you believe that, it follows that the intersection of the conjugates of $SL_n(\mathbb{Z})$ in $SL_n(\mathbb{R})$ is either $\pm I$ or trivial, and so $SL_n(\mathbb{Z})$ is very far from normal.