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For a normal subgroup $H⊴G$, the quotient group $G/H$ is the set of cosets $gH$ with multiplication $(gH)(kH)=gkH$. If $gH=g^{\prime }H$ and $kH=k^{\prime }H$, 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 $\mathrm{ord}(gH)$ 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 $\phi (gh{g}^{-1})=\phi (g)\phi (h)\phi (g{)}^{-1}=1$ whenever $h\in \ker \phi$. Conversely, if $H$ is normal, the quotient map $G\to G/H$ has kernel $H$. Thus $H⊴G$ exactly when it is the kernel of a group homomorphism.

A group is 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 areAll 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