= Solution
For a <normal subgroup> $H\trianglelefteq G$, the <quotient group> $G/H$ is the set of cosets $gH$ with multiplication $(gH)(kH)=gkH$. If $gH=g'H$ and $kH=k'H$, normality moves the intervening elements of $H$ past $k$, proving that the product is independent of representatives.
If $g$ has finite order $m$, then $(gH)^m=H$, so $\operatorname{ord}(gH)$ divides $m$. For finite $G$, the greatest element order in $G/H$ is therefore no greater than that in $G$.
Every kernel is normal because $\varphi(ghg^{-1})=\varphi(g)\varphi(h)\varphi(g)^{-1}=1$ whenever $h\in\ker\varphi$. Conversely, if $H$ is normal, the quotient map $G\to G/H$ has kernel $H$. Thus \b[$H\trianglelefteq G$ exactly when it is the kernel of a group homomorphism].
A group is <metacyclic group>[metacyclic] when it has a cyclic normal subgroup with cyclic quotient. A <dihedral group> has its cyclic rotation subgroup normal and quotient $C_2$, so every dihedral group is metacyclic.
Up to isomorphism the groups of order eight are
$$
C_8,\quad C_4\times C_2,\quad C_2^3,\quad D_8,\quad Q_8.
$$
All except $C_2^3$ are metacyclic: use a cyclic subgroup of order four in the three noncyclic examples that have one. In $C_2^3$, every cyclic subgroup has order at most two and its quotient is not cyclic. The normal subgroups of $A_4$ are $1,V_4,A_4$; none gives the required cyclic subgroup and quotient, so $A_4$ is not metacyclic. Finally, $S_1,S_2,S_3$ are metacyclic, while $S_4$ and $S_5$ are not, by their normal-subgroup structures. Hence
$$
\boxed{S_n\text{ is metacyclic for }n\leq5\text{ exactly when }n=1,2,3.}
$$
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