My grammar can be awkward because my native language is not English. Please excuse me.
[Problem]
There are $ \alpha, \beta\in K$ that is a field extension of $ \mathbb{F}_{p}$ such that $[\mathbb{F}_{p}(\alpha): \mathbb{F}_{p}]=12,[\mathbb{F}_{p}(\beta): \mathbb{F}_{p}]=20 $.
Show that $[\mathbb{F}_{p}(\alpha,\beta): \mathbb{F}_{p}]=60 $ and $[\mathbb{F}_{p}(\alpha)∩\mathbb{F}_{p}(\beta): \mathbb{F}_{p}]=4$.
Let $ G(\mathbb{F}_{p}(\alpha,\beta)/\mathbb{F}_{p})=(\phi)$ where $ \phi$ : $\mathbb{F}_{p}(\alpha,\beta)\to\mathbb{F}_{p}(\alpha,\beta)$, $ \phi(u)=u^{p}$ and order of $ {\phi}$ be $n$.
Then there exists $d|n$ such that $G(\mathbb{F}_{p}(\alpha)/\mathbb{F}_{p})=(\phi^{d})$.
so $12=[\mathbb{F}_{p}(\alpha):\mathbb{F}_{p}]=[(\phi):(\phi^{d})]=\frac{n}{n/d}=d$.
Similarly we can show that $G(\mathbb{F}_{p}(\beta)/\mathbb{F}_{p})=(\phi^{20})$.
I think that $G(\mathbb{F}_{p}(\alpha)∩\mathbb{F}_{p}(\beta)/\mathbb{F}_{p})$ is smallest group containing $(\phi^{12})$ and $ (\phi^{20})$, that is $(\phi^{4})$ , and $G(\mathbb{F}_{p}(\alpha,\beta)/\mathbb{F}_{p})=(\phi^{12})∩(\phi^{20})=(\phi^{60})$, but I don't know the exact reason.
Further on if $[\mathbb{F}_{p}(\alpha): \mathbb{F}_{p}]=m,[\mathbb{F}_{p}(\beta): \mathbb{F}_{p}]=n $. Then $[\mathbb{F}_{p}(\alpha,\beta): \mathbb{F}_{p}]=\text{lcm}(m,n) $ and $[\mathbb{F}_{p}(\alpha)∩\mathbb{F}_{p}(\beta): \mathbb{F}_{p}]=\gcd(m,n)$. Is it correct?