Finite cyclic group

A finite cyclic group of order $n$ is isomorphic to the additive group $C_n=\mathbb{Z}/n\mathbb{Z}$. The group isomorphism sends the congruence class of $k$ to $g^k$, where $g$ is a chosen generator of a group. For $n=6$, the identity element is $0$, $4+5=3$ modulo six, and the inverse element of $4$ is $2$.

One finite cyclic group can have several choices of generator of a group. Adding $1$ visits all six elements before returning to zero; adding $5$ also visits all six, in the reverse order. Adding $2$ reaches only $\{0,2,4\}$ from zero, and adding $3$ reaches only $\{0,3\}$. The following permutations of the six elements show what each repeated step can reach.

This gives a precise version of a familiar scheduling question: does a fixed step visit every position, or repeatedly miss some? The generators of a finite cyclic group are exactly the steps coprime to the total number of positions. Changing the step changes the orbit through the same underlying group; the diagram does not replace its group operation with a different one.

The cyclic group generated by $r$ with $r^4=e$ and $r^2\ne e$ has elements $e,r,r^2,r^3$. It is realized by quarter-turn rotations of a plane. The cyclic group of order four and Klein four-group have the same order but different element orders: the latter has no element of order four.

For real $b,c$, this matrix defines a group representation of $C_4=⟨a:a^4=e⟩$ exactly when $bc=\pm 1$. Indeed its lower block $B$ satisfies $B^2=bc{I}_2$, so $T(a{)}^4=I_3$ exactly when $(bc{)}^2=1$. If $bc=-1$, $T(a{)}^2=\mathrm{diag}(1,-1,-1)\ne I_3$, so its order is four and the representation is faithful. If $bc=1$, its order is exactly two and the kernel of a group homomorphism is $\{e,a^2\}$, giving an unfaithful representation. Other values do not give representations at all, including $bc=0$, when the proposed matrix is singular.

For a positive integer $n$ and a congruence class $k$ in the additive finite cyclic group $\mathbb{Z}/n\mathbb{Z}$, the order of a group element isIndeed, write $d=\gcd (n,k)$, $n=d{n}^{\prime }$ and $k=d{k}^{\prime }$. Returning to zero after $m$ steps means $n\mid mk$, equivalently $n^{\prime }\mid m{k}^{\prime }$. Since $n^{\prime }$ and $k^{\prime }$ are coprime, this holds exactly when $n^{\prime }\mid m$; the least positive such $m$ is $n^{\prime }$. This argument also covers $k=0$, since $\gcd (n,0)=n$.

For $n=6$, the orders of $0,1,2,3,4,5$ are respectively $1,6,3,2,3,6$. The translation in a group $x\mapsto x+k$ has the same period from every starting element. Consequently its cycles each have length $n/d$ and there are $d$ of them. They are exactly the cosets of the cyclic subgroup generated by $k$, which explains both the lengths and the number of components in the picture.

For any group $G$, evaluation at a chosen generator of a group $x$ of $C_n$ gives a bijection between group homomorphisms $C_n\to G$ and elements satisfying $g^n=1$. A group homomorphism must send $x^j$ to $g^j$. Conversely, this formula is well-defined because exponents differing by a multiple of $n$ give equal powers, and it respects multiplication. The displayed correspondence is a bijection of sets; it is not claimed to be a group isomorphism for a nonabelian target. In a symmetric group, the order of a group element is the least common multiple of its disjoint cycle lengths, making this correspondence easy to enumerate.