Are cyclic groups defined for only two operations (multiplication, addition)?

Question

I am learning group theory now I’m on the topic cyclic groups.

Textbooks say that

a group is cyclic if it is generated by an element $a$ of this group,

or

a group $G$ is said to be cyclic if every element of $G$ is a power of one and the same element $a$ (say) of $G$ and written as $$G=\langle a : a^n = e \rangle$$ (read as $G$ is a cyclic group of order $n$ generated by $a$).

My question is that is $$ a^n = \underbrace{a \cdot a \cdot a \dotsm a}_{\text{$n$ times}} $$ operation is multiplication only? Or it can be written for any operation $*$?

I’m confused because every text book I read gives examples under multiplication operation.

Can you help me to give examples of cyclic groups which are cyclic under operation different from multiplication or addition?

Thank you in advance.

Answer 1

A group is in some sense a generalization of some familiar concepts, including familiar types of addition and multiplication.

The way of denoting the group operation is arbitrary, though sometimes suggestive. For instance addition, $+$, is often used for abelian groups.

The group operation can be any binary operation that satisfies the requirements for a group.

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Any element $a$ of any group $G$ generates a cyclic subgroup. If the order of $a$ is finite (or $a$ is torsion), say equal to $n$, then $\langle a\rangle \cong \Bbb Z_n$ (or $C_n$). If $a$ is not torsion, then $\langle a\rangle \cong \Bbb Z$.

Consider as an example the dihedral group $D_{2n}$, generated by a rotation of order $n$ and a reflection of order $2$ (and subject to another anti-commutativity relation). Suggestive notation here could be composition, $\circ$. Or to be perfectly frank, one might as well just use $•$ for any group (the point is it doesn't matter).

Now the rotation $r$ generates a cyclic subgroup of order $n$. And the generator $s$ a (cyclic) group of order $2$. You'll learn more about this group as you go along.

Answer 2

Each group $\mathcal{G}=(G, \cdot, e)$ is a set $G$ under a binary operation $\cdot: G\times G\to G$ with a distinguished element $e\in G$ under the axioms:

1. Closure: For all $a,b\in G$, $a\cdot b\in G$. (This is often omitted, since, really, it follows from the definition of a binary operation.)

2. Associativity: For all $a,b,c\in G$, $$a\cdot (b\cdot c)=(a\cdot b)\cdot c.$$

3. Identity: For all $a\in G$, $$a\cdot e=a=e\cdot a.$$

4. Inverses: For all $a\in G$, there exists an $a^{-1}\in G$ such that $$a\cdot a^{-1}=e=a^{-1}\cdot a.$$

Often, we drop mentioning $\cdot$ and use concatenation instead, thereby writing $a\cdot b$ as $ab$. That is what happens in your question.

Each (finite) cyclic group $H$ is isomorphic to a group $\mathcal G=(G,\cdot, e)$ such that $G=\langle g\mid g^n=e\rangle $ for some $n\in \Bbb N\cup\{0\}.$ By this, we mean that there exists a bijection $\varphi: G\to H$ such that for all $a,b\in G$, we have $$\varphi(a\cdot b)=\varphi(a)\varphi(b).$$

In summary: your $\ast$ is not necessarily multiplication or addition, nor is your $\cdot$; rather, they are arbitrary, abstract symbols.

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Note that there is nothing special about groups here. Given any set $S$ with an associative binary operation $\star: S\times S\to S$, we write

$$s^n=\underbrace{s\star\dots\star s}_{n\text{ times}}$$

or even

$$ns=\underbrace{s\star\dots\star s}_{n\text{ times}}$$

for any $s\in S$.