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5

Start with the $n$ dimensional irrep of $\mathfrak{sl}_2$ and restrict it to its Borel subalgebra of upper triangular matrices (a non-abelian $2$ dimensional Lie algebra). It is indecomposable because any proper submodule does not contain the highest weight vector. This produces one indecomposable in every dimension.


2

Supplementing Stephen's excellent answer (+1) with concrete matrices to cater for the possibility that this exercise was encountered prior to the theory of representations of $\mathfrak{sl}_2$. Let us fix a dimension $n$. Let $A$ be the diagonal matrix with entries $-(n-1)/2$, $-(n-3)/2$, $\ldots$, $(n-3)/2$,$(n-1)/2$ in an arithmetic progression from ...


2

No, that would be saying that every representation is completely reducible to a direct sum of 1-dimensional representations. For instance, $S_3$ the symmetric group on 3 letters has a simple (thus already reduced) representation on $\mathbb{C}^2$ where a 3-cycle $\tau$ acts as rotation through $2\pi/3$ and a 2-cycle $\sigma$ acts as a reflection. Now ...


2

No. For finite groups, the basic problem (as observed in the comments) is that a representation induced from a subgroup $H$ of a finite group $G$ has dimension divisible by $[G : H]$. In particular, if $H$ is a proper subgroup then an induced representation can't have dimension $1$, but there also exist finite groups $G$ such that the smallest index of a ...


1

If you are proving by contradiction, as it seems you are, then we must show that the existence of a $1$-dimensional invariant subspace leads to a contradiction. Any $1$-dimensional subspace $U$ in $\mathbb{R}^2$ is spanned by a unit vector $u$. Now consider the action of any $r\in \mathbb{R}$ on $u$ that is not a multiple of $\pi$. You should find that ...


1

The elements of $\mathfrak{sp}_4(\mathbb{C})$ can be written in block form $$ X=\begin{pmatrix} A & B \\ C & -A^t \end{pmatrix} $$ with $2\times 2$ blocks, where $B^t=B$ and $C^t=C$. From here we count that the dimension of the Lie algebra equals $10$, because $A$ has $4$ parameters, and $B$ and $C$ have three parameters (so that $4+3+3=10$). Now ...


1

A partial answer is the image of a faithful representation $\mathbb{V}$ of a group (endowed with the group operation of composition, or in a basis, matrix multiplication) is isomorphic to the group itself, and hence realizes your group as a subgroup of $GL(\mathbb{V})$ (or given a basis, as an explicit matrix group), whereas the image of a nonfaithful ...


1

Any normal subgroup, not only the commutator subgroup, Is the union of (disjoint) conjugacy classes.


1

Here is an answer derived from the answer of Stephen. First, write the equivalent relation $[ - e_2, e_1]= e_1$ and denote $e = - e_2$, $f= e_1$ to get $[e,f] = f$. For $\lambda$ a parameter ( take it $\text{mod}\ \mathbb{Z}$) consider the infinite dimensional vector space $V_\lambda$ spanned by $v_i$, $i \in \mathbb{Z}$. It is easy to check that the ...


1

This is for part $2$. First, we have the following $A^+$-submodules of $F^n$ $$V_k = \textrm{span}\{ e_i \ | \ 1\le i \le k\}$$ for all $0\le k \le n$, with $V_0 = (0)$, $V_n = F^n$ and $\dim V_k = k$ for all $k$. We will prove there are no other $A^+$ submodules. Consider the following element of $A^+$, $g= I +n$, where $n(e_i) = e_{i-1}$, $n( e_1) = ...


1

in GAP we get: G:=Group((1,2)(5,6),(2,4)(3,5),(2,6)(3,7));; StructureDescription(G); "PSL(3,2)" A group can be generated by a different number of generators. The symmetric group $S_n$ is generated by a cyclic permutation of order $n$ and a permutation of two elements (e.g. $(1..n)$ and $(1,2)$). In this case there are only two generators, but the group ...



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