TITLE: Irreducible representations over $\Bbb R$
QUESTION [5 upvotes]: How to prove that all irreducible representations over $\mathbb{R}$ of finite abelian group have dimension 1 or 2?

REPLY [5 votes]: For all elements $g$ in your finite abelian group $G$, you have an endomorphism on your vector space $V$ underlying your representation. Since your group is abelian, these endomorphisms commute, hence you can diagonalize them simultaneously (over $\mathbb C$). This proves, of course, that over the complex numbers (or any algebraically closed field), your irreducible representations are one-dimensional.
Over $\mathbb R$, the best you can hope for is that your endomorphisms can be block-diagonalized simultaneously, with blocks of size at most 2. This is because an irreducible polynomial over $\mathbb R$ has degree at most 2. EDIT: Let's prove this. If your polynomial has odd degree, then it has a root (this is a consequence of the intermediate value theorem). Now, suppose your polynomial has even degree. By looking at its solutions over $\mathbb C$, we know they come in conjugate pairs $z$, $\overline z$. Therefore, both $(X-z)$ and $(X-\overline z)$ divide your polynomial, and therefore, so does their product, which is a quadratic polynomial with real coefficients. Therefore, any real polynomial of degree at least 3 is reducible.
Therefore, your irreducible representations are at most two-dimensional.