Let $\alpha \in \mathbb{R}$, let $\mathbb{T}$ be the set of complex numbers with absolute value $1$ and let $\theta:\mathbb{R} \to \mathbb{T} \times \mathbb{T}, \; \theta(t)=(e^{it}, e^{\alpha i t})$ be a group homomorphism.
Show that $\ker\theta$ is non-trivial if and only if $\alpha \in \mathbb{Q}$.
Now $\ker\theta = \{x \in \mathbb{R} : (e^{ix}, e^{\alpha i x})=(1,1)\}$ and I have the implication that $\alpha \in \mathbb{Q} \implies \ker\theta$ is non-trivial, but I can't express a closed argument for the reverse implication.
Please help!
Suppose $t \neq 0$ is in the kernel of $\theta$. Then $e^{it}$ and $e^{\alpha it}$ are both $1$, so $t$ and $\alpha t$ are both integer multiples of $2\pi$.
If $t = 2\pi m$ and $\alpha t = 2\pi n$ for integers $m$ and $n$, then by dividing the two (remember $t \neq 0$), $\alpha= \frac{n}{m}$.