Exercise 4, § 6

(a) Let G be a finite group of order $mn$ such that there exists a normal subgroup H of G which is cyclic of order m, the quotient group G/H being cyclic of order n. Show that G is generated by two elements a, b such that $a^m = e, b^n = a^r, bab^{-1} = a^s$, where r and s are two integers such that $r(s - 1)$ and $s^n - 1$ are multiples of m (take a to be an element generating H, b an element of a coset generating G/H; express the elements $b^h a^k b^{-h}$ as powers of a and apply this in particular to the cases $h = n, k = 1$ and $h = 1, k = r$).

(b) Conversely, let $G(m, n, r, s)$ be the group defined by the presentation

where m and n are two integers $\geqslant 0$ and r and s arbitrary integers (cf. § 7, no. 6); show that, if m, $r(s - 1)$ and $s^n - 1$ are not all zero, $G(m, n, r, s)$ is a finite group of order $qn$, where q is the greatest common divisor of m, $|r(s - 1)|$ and $|s^n - 1|$; in this group, the subgroup H generated by a is a normal subgroup of order q and G/H is a cyclic group of order n (prove that every element of $G(m, n, r, s)$ can be written in the form $a^x b^y$, where x and y are two integers such that $0 \leqslant x \leqslant q - 1, 0 \leqslant y \leqslant n - 1$, and $G(m, n, r, s)$ is isomorphic to the group consisting of the ordered pairs $(x, y)$ of integers subjected to the above conditions, with law of composition

the first coordinate of the right-hand side being a sum modulo $q$. Examine the cases where $m = r(s - 1) = s^n - 1 = 0$.

The group $G(n, 2, 0, -1)$ is called the dihedral group of order $2n$ and is denoted by $\mathbf{D}_n$; the group $G(4, 2, 2, -1)$ is a group of order 8 called the quaternionic group and denoted by $\mathfrak{Q}$. Show that in $\mathfrak{Q}$ every subgroup is normal and that the intersection of the subgroups distinct from $\{e\}$ is a subgroup distinct from $\{e\}$. Prove that $\mathbf{D}_4$ is not isomorphic to $\mathfrak{Q}$*