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Given a cyclic-factor subnormal series $1=G_0⊴\cdots ⊴G_n=G$ and $H\le G$, intersect each term with $H$. The induced map $(H\cap G_{i+1})/(H\cap G_i)\to G_{i+1}/G_i$ is injective. Since every subgroup of a cyclic group is cyclic, the intersection chain is again a cyclic-factor subnormal series, after removing repetitions. Hence $H$ is a polycyclic group. In particular, every subgroup is finitely generated.

A subnormal series is a finite chain of subgroups in which each is a normal subgroup of the next. A member need not be normal in the whole group. Requiring the factors to be cyclic groups defines a polycyclic group; allowing finite factors as well defines a poly-(cyclic or finite) group.

For finite sets $A,B\subseteq G$ in an abelian group $G$, their sumset isEach distinct value occurs only once in this set, even if many ordered pairs produce it. When $B=A$, repetitions are allowed: $a+a$ belongs to $A+A$. For the two integer sets introduced above,For $A$, the value $3$ comes from $(0,3),(1,2),(2,1),(3,0)$. For $B$, two ordered pairs give the same sum only when one reverses the other, apart from a pair being identical to itself. Thus its four pairs $(b,b)$ and six pairs with distinct members, counted without regard to order, produce ten distinct sums.

Every nonempty finite set of integers $S\subseteq \mathbb{Z}$ satisfies $|S+S|\ge 2|S|-1$. To see this, list its members as $s_1<\cdots <s_n$. The $2n-1$ distinct sumsall belong to $S+S$. A nonconstant arithmetic progression attains this bound. The ordered list is essential: in the cyclic group $\mathbb{Z}/6\mathbb{Z}$, the subgroup $H=\{0,2,4\}$ instead satisfies $H+H=H$, of cardinality three. Addition wraps around, for example $4+4=2$ in this group.

For an abelian group $G$ and an integer $m\ge 1$, its $m$-torsion subgroup is $G[m]=\{x\in G:mx=0\}$, the kernel of multiplication by $m$. In a finite abelian group, every nonempty fiber of this homomorphism has cardinality $|G[m]|$, so $|G:mG|=|G[m]|$. This follows directly: if $mx=my$, then $x-y\in G[m]$, and conversely translating by an element of $G[m]$ leaves the image unchanged. On the cyclic group ${\mathbb{Z}}_N$, the solutions of $mx=0$ number $\gcd (m,N)$; divide both integers by their greatest common divisor to see that $x$ must be a multiple of $N/\gcd (m,N)$.

A group action on a nonempty set $X$ is transitive when every two points lie in the same orbit of a group action, equivalently when for all $x,y\in X$ there is a group element $g$ with $g\cdot x=y$. There is exactly one group orbit: every position is reachable from every other. Requiring $X\ne \varnothing$ avoids calling the group action on the empty set transitive merely because the quantified statement has no points to test.^[ref]

Here a left group action of a group $G$ means a function $G\times X\to X$, written $(g,x)\mapsto g\cdot x$, satisfyingwhere $e$ is the identity element. Products move the point by $h$ first, then by $g$. Transitivity says that an orbit map $G\to X$, $g\mapsto g\cdot x$, is surjective for every $x\in X$; it suffices to check one chosen point.^[ref][ref]

For example, the symmetric group $S_3$ acts on $X=\{1,2,3\}$ by $\sigma \cdot i=\sigma (i)$. This is a transitive group action, but it is not a free group action: the stabilizer subgroup of $1$ is $\{e,(23)\}$. Both the transposition $(12)$ and the cyclic permutation $(123)$ send $1$ to $2$. In contrast, the cyclic group $C_3=\mathbb{Z}/3\mathbb{Z}$ acts on itself by $k\cdot x=k+x(\mathrm{mod}3)$, using modular arithmetic. The group element sending $x$ to $y$ is uniquely $k=y-x(\mathrm{mod}3)$. These two group actions reach the same number of points, but only the second has unique transport between them.

The stabilizer subgroup $G_x=\{h\in G:h\cdot x=x\}$ measures the ambiguity of transport. If $g_0\cdot x=y$, thenThe first set is a left coset, and the second identity says that stabilizer subgroups in one group orbit are conjugate subgroups. Briefly, $g\cdot x=g_0\cdot x$ exactly when $g_0^{-1}g\in G_x$. Thus, for a finite group in a transitive group action, the orbit-stabilizer theorem gives $|G|=|X||G_x|$, and there are exactly $|G_x|$ transports for each pair of points. The example counts are $6=3\cdot 2$ for $S_3$ and $3=3\cdot 1$ for $C_3$. This separates the number of reachable positions from the number of ways to reach each one.^[ref]

Embed $A\subseteq [N]$ in the odd cyclic group $G=\mathbb{Z}/(2N+1)\mathbb{Z}$, so additive three-term relations in $A$ cannot wrap around. Form three copies of $G$, putting edges $x\sim y$ when $y-x\in A$, $y\sim z$ when $z-y\in A$, and $x\sim z$ when $(z-x)/2\in A$. A triangle in a graph encodes $a+b=2c$ in $A$. If $A$ has no nonconstant arithmetic progression, every triangle in a graph has $a=b=c$; there are exactly $|G||A|$ triangles in a graph, and they are edge-disjoint. Thus at least $|G||A|$ edges must be removed to eliminate them. For fixed positive subset density of $A$, this is a positive fraction of the squared vertex count, whereas the triangle in a graph count is only quadratic. The triangle removal lemma permits deletion of an arbitrarily small quadratic fraction when the cubic triangle in a graph subset density is sufficiently small. This contradiction establishes Roth theorem on three-term arithmetic progressions.

A finite abelian group is a cyclic group if and only if it has at most one subgroup of each order dividing its order. A cyclic group $C_n$ has exactly one subgroup of order $d$ for each $d\mid n$, generated by the $n/d$th power of a generator.

Conversely, the structure theorem for finitely generated modules over a principal ideal domain expresses $A$ as $C_{d_1}\oplus \cdots \oplus C_{d_s}$ with $2\le d_1\mid \cdots \mid d_s$. If $s\ge 2$, choose a prime number $p\mid d_1$. The first and second summands then provide distinct subgroups of order $p$, contradicting the hypothesis. Hence $s\le 1$, so $A$ is cyclic, including the trivial group when $s=0$. This criterion combines with the root bound for a polynomial to prove that a finite subgroup of a field multiplicative group is cyclic.