In Fulton and Harris' "Representation theory, A first course" on page 150 they state
Note also by the symmetry of the eigenvalues we may deduce the useful fact that any representation $V$ of $\mathfrak{sl}_{2}(\mathbb{C})$ such that the eigenvalues of $H$ all have the same parity and occur with multiplicity one is necessarily irreducible.
I would like to know why this is true. I appologise in advance if this is obvious but I am not really seeing why this is so.
Note that $H=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$, one of the elements in the standard basis of $\mathfrak{sl}_{2}(\mathbb{C})$. Though I have found this chapter of Fulton and Harris confusing I have read this http://math.uchicago.edu/~may/REU2012/REUPapers/Seitz-McLeese.pdf and understand this a lot better. I understand that up to isomorphism there is one irreducible representation of dimension $n$ of $\mathfrak{sl}_{2}(\mathbb{C})$.
Let $V$ be a $\mathfrak{sl}_2(\mathbb{C})$-module such that all eigenvalues of $H$
If $V$ was not irreducible, it would have a minimal non-trival sub-module $W$ (since $V$ is finite-dimensional), which would therefore be irreducible. So, the action of $H$ on $W$ would be diagonalizable (they prove it on p. 147). Let $w\in W$ be a vector with a maximal eigenvalue $m$ with respect to the action of $H$ on $W$. Then $X.w=0$, since $H.(X.w)=(m+2).(X.w)$ and $m+2>m$. On the other hand, if $v\in V$ is a vector with a maximal eigenvalue $n$ with respect to the action of $H$ on $V$, then $n>m$ (if they were equal, $W=V$) and $n$ and $m$ have the same parity. Then both $w$ and $Y^{\frac{n-m}2}.v$ are eigenvectors (of the action of $H$) with eigenvalue $m$. Since the space of such eigenvalues is $1$-dimendional, they are multiples of each other. But this is impossible, because $X.(Y^{\frac{n-m}2}.v)\neq0$.