What is a group and its operation?

Question

I am learning group theory by myself and has not reached cyclic groups yet but as I have read cyclic groups are group generated by a single element and is denoted as $G = \langle x, *\rangle$ where $*$ is a operation.

I have some questions:

1. Can this $*$ be a function? I think yes it can be a function. Because every operation is a kind of function.

If yes, then will the following is a group, specifically cyclic group:

2. $G = \langle x, 26(x^2+8x) \mod 103\rangle$ where $x = 2$. I think yes and no (I am confused). Yes because it generates a cyclic group of order $9$, $G = [2, 5, 42, 10, 45, 4, 12, 60, 93]$. No because the operator here is a unary operation that would not be able to combine two elements of group.

I might have made an error in notation so please ask in comments for any clarification.

Answer 1

Your confusion seems to be about the meaning of the phrase "generated by".

• with its informal meaning, it is true that starting from the element $2\bmod{103}$, the function $f:\Bbb Z/(103\Bbb Z)\to \Bbb Z/(103\Bbb Z),\;x\mapsto26(x^2+8x)$ "generates" the subset $G:=\{2, 5, 42, 10, 45, 4, 12, 60, 93\}$, more precisely: $f(2)=5,f(5)=42,\dots,f(60)=93,f(93)=5$, and then the sequence repeats periodically.
  But as long as you don't equip this set $G$ with some binary operation (i.e. some function from $G\times G$ to $G$), you have no opportunity to claim it to be (or not to be) a group.
• with the formal and specific meaning of group theory, one generally speaks of a group $G$ generated by a subset $S\subseteq G$. The group $G$ here is supposed to be already given, implicitely equipped with its binary operation. In the special case where $S$ is a singleton $\{s\}$, one says that $G$ is generated by $S$ or simply by $s$, if every element of $G$ is of the form $s^k$ (if the binary operation of the group is denoted mutiplicatively, $\cdot$) or $ks$ (if it is denoted additively, $+$) for some integer $k$. For instance the group $(\Bbb Z,+)$ is generated by $1$ and the group $(\{1,i,-1,-i\},\cdot)$ is generated by $i$.

Answer 2

A superfast overview:

there's only one cyclic group of order $n$, and the prototype is $(\Bbb Z_n,+)$ with addition. The integers modulo $n$. Addition is used because the group is abelian (commutative).

The infinite cyclic group is the integers $(\Bbb Z,+)$.

Those are all the cyclic groups.

Answer 3

Each group requires a binary operation, which, yes, is a kind of function. Each binary operation takes two things (that may be duplicates) as input to work, not one.

Definition: A group $(G,\circ)$ is a set $G$ under a binary operation

$$\circ:G\times G\to G$$

that is closed (by definition of a binary operation), associative, has an identity element $e\in G$, and each element of $G$ has an inverse in $G$ with respect to $e$.

Your candidate group does not have a binary operation, so cannot be a group.

To say that the set of integers you list is a group is mistaken. Remember, you need, in addition to the set, the binary operation; and the same set can produce different groups under different binary operations.

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It looks like you are trying to run before you can walk. Try a few more basic exercises in group theory ahead of tackling cyclic groups.