Q1) Prove that $G$ contains an element of order 20
Q2) Assume $\exists H\subset G$ s.t $H$ normal in $G$ and |$\phi(H)$|=60. Prove that $G$ contains a normal subgroup $K$ such that |$G/K$|=36
For part a), I showed that $\exists$$a$$\in$$G$ s.t |$a$|=n and |$\phi(a)$|=20, thus 20|n. How do i complete this part?
For b), I am at a complete roadblock.
$S_4\times D_{15}$ contains an element $y$ of order $20$ because $S_4$ contains an element of order $4$ and $D_{15}$ an element of order $5$. By surjectivity of $\phi$, let $a\in G$ with $\phi(a)=y$. Since $|G|$ is finite, $a$ has finite order, $n$ say. Then $1=\phi(a^n)=y^n$ implies $20\mid n$, so $n=20k$ for some $k\ge1$. Then $a^k$ has order $20$.
$S_4$ as a normal subgroup of order $4$ (isomorphic to the Klein four group), in other words, there is an epimorphism $S_4\to S_3$. We also have an epimorphism $D_{15}\to D_3$ ("raising to fifth power"). This gives us an epimorphism $G\to S_3\times D_3$. The kernel $K$ of this epimorphism has index $|S_3\times D_3|=36$ in $G$.