Isomorphism between reducible subgroup and irreducible subgroup? Let $ H_1 $ be a closed group of matrices acting irreducibly on $ \mathbb{C}^n $ and let $ H_2 $ be a closed group of matrices acting reducibly on $ \mathbb{C}^n $. Is it possible for the groups $ H_1,H_2 $ to be isomorphic?
For $ n=1 $ this is impossible since every matrix group acts irreducibly.
For $ n=2 $ this is impossible because
Claim: A subgroup of $ GL_2(\mathbb{C}) $ is reducible if and only if it is abelian.
The reverse direction for this claim follows from the fact that commuting operators can be simultaneously diagonalized and the forward direction follows from the fact that for $ n=2 $ every group acting reducibly is conjugate to a group of diagonal matrices.
Since two isomorphic groups are either both abelian or bot not abelian then it is impossible to have a pair of subgroups in $ GL_2(\mathbb{C}) $ one of which is abelian and one is not.
 A: Yes it is possible, but you need to go up to dimension 3.
There are two ways to embed $SL(2, \mathbb{C})$ into $GL(3, \mathbb{C})$, one where it acts reducibly and one where it acts irreducibly.
When I say 2 I mean upto conjugation, in reality there are of course infinitely many ways.
The reducible case is easy to guess: $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
The irreducible way I will describe in words and leave writing down the explicit morphism w.r.t. a suitable basis (or to put it more in the wording of the question: to find the group $H_1$ which is the image of this morphism) to you:
Identify the the vectorspace $\mathbb{C}^3$ with the three dimensional subspace $$\{\begin{pmatrix} a & b \\ c & -a \end{pmatrix} \colon a, b, c \in \mathbb{C} \} \subset Mat(2, \mathbb{C})$$
of traceless two by two matrices.
Each $G \in SL(2, \mathbb{C})$ acts on this space by $X \mapsto GXG^{-1}$.
(I leave it to you to verify that $GXG^{-1}$ is indeed traceless again).
This action is known as the adjoint action of $SL(2, \mathbb{C})$ on its own Lie algebra.
This action is irreducible, which of course also needs verification, although it is somewhat clear intuitively. Let me know if you need help with that part.
EDIT: as per the comments I will say something about why the second action is irreducible.
First of all we note that for $X \neq 0$ to live in a one-dimensional invariant subspace, requires $X$ to commute with all other 2 by 2 matices You can check that that means that $X$ is a scalar multiple of the identity. But the requirement that $X$ has trace 0 then implies that $X = 0$, ruling out this case.
The only thing to worry about then are two-dimensional invariant subspaces. (Of course any trick for ruling them out will also rule out the one-dimensional case, so the above remark is unnecessary, but I leave it in because it is so nice.)
A very nice thing to recognize about this action is that it is also what we use to bring matrices in Jordan normal form. That is: if $J$ is the normal form of $X$ then $J = GXG^{-1}$ for some $G$ in $SL(2, \mathbb{C})$.
Hence if $X$ is part of some invariant subspace $V$, then so is its normal form $J$. Since moreover $X$ is traceless we know what this form looks like: either
$$
\begin{pmatrix}
\lambda & 0 \\
0 & - \lambda
\end{pmatrix}
$$
or
$$
\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix}
$$
These matrices are so nice, we might as well start over: suppose that an invariant subspace $V$ contains a matrix $J$ from this form. Then we can show (pretty much by brute force) that there exist $G_1, G_2 \in SL(2, \mathbb{C})$ such that the three matrices $J, G_1JG_1^{-1}, G_2JG_2^{-1}$ are all linearly independent from each other. Since they all must live in the subspace $V$, we see that $V$ is at least 3-dimensional and hence equal to the full space of all traceless matrices.
