What is a cyclic group? I cannot seem to wrap my head around cyclic groups. I need this explained in layman's terms because this seems to be a big part of Modern Algebra, and its confusing me with other stuff. I know that,
a group is cyclic if there exist an $a \in G$ such that every element of $G$ is a power of $a$. We write $G=\langle a \rangle$ or $a$ is a generator of $G$.
So for example, would $\mathbb{Z}_6$ be a cyclic group? $\mathbb{Z}$ is a group under addition, so it has generators $\langle 1 \rangle = \langle 5 \rangle$. So, would that make $\langle1 \rangle=\langle 5 \rangle$ cyclic subgroups, therefore $\mathbb{Z}_6$ is a cyclic group?
 A: We are going to make it geometric.
Let's first cover finite cyclic groups.  A cyclic group of finite order $N$ is the group of rotations, in the $xy$-plane, about the origin by the following angles:
$$
{2 \pi k \over N}, \quad k = 0, 1, 2, \ldots, N-1.
$$
In such a group, if we take the (smallest nontrivial) rotation, $k=1$, which rotates the plane by the angle
$$
{2 \pi \over N},
$$
all other rotations in the group can be obtained by repeating this one.  That's how this rotation generates all the others.  This rotation is general not the only generator.
Sketch the rotations for $N=4$: these rotate the plane through quarters of a full turn.  Study what they do to a square with vertices $(1,1), (-1,1), (-1,-1), (1,-1)$.  (The "preserve" this square.)
Now return to your example: $N=6$.  What polygons are preserved by the rotations in this cyclic group?  Try multiples of rotation $k=5$.  Do these multiples give us the entire group?  Why about the multiples of rotation $k=3$?
And if the cyclic group is infinite, it is isomorphic to $\mathbb{Z}$.  It shifts the real line by multiples of integers.  What generators does this group have?
A: You're absolutely right! A group is cyclic if it has an element that generates the group. In the case of $\Bbb Z_6,$ you've found the two possible generators, though we should probably think in terms of multiples (vs. powers), since we're dealing with commutative groups.
Another way to think about it--which gets around distinctions like multiples vs. powers--is this:


*

*Given a group $G,$ let $\mathscr{S}(G)$ indicate the set of subgroups of $G$. For example, $G$ and the trivial subgroup (with just the identity of $G$) will be elements of $\mathscr{S}(G).$

*Given any set $\mathscr A\subseteq \mathscr S(G),$ if $\mathscr A$ is non-empty, then we can prove that $\bigcap\mathscr A\in\mathscr S(G)$ using the subgroup test.

*Given any subset $A$ of $G,$ we define $\langle A\rangle:=\bigcap\bigl\{S\in\mathscr{S}(G):A\subseteq S\bigr\}.$ By the above, $\langle A\rangle$ is the smallest subgroup of $G$ containing $A$ as a subset.

*When considering sets of the form $\{a\}$ for $a\in G,$ we will typically use $\langle  a\rangle$ to mean the same thing as $\bigl\langle\{a\}\bigr\rangle,$ just because it looks nicer. Similarly, we'll do the same thing for other finite sets. From that perspective, saying that there is some $a$ such that $\langle a\rangle=G$ means that no proper subgroup of $G$ has $a$ as an element, because the whole group $G$ is the closure of $\{a\}$ under the group operation, inverses, and identity.

A: Actually a cyclic group is isomorphic to $\mathbb{Z}/n\mathbb{Z}$ for some $n=1,2\dots$ or it is isomorphic to $\mathbb{Z}$(you may think this as the case $n=0$).
  Note that $(\mathbb{Z}/n\mathbb{Z},+)$ is a cyclic group of order $n$, since it is generated by $\overline{1}$. As you may have noticed, this generator is not unique, in fact $(\mathbb{Z}/n\mathbb{Z},+)$ has $\varphi(n)$ generators. For instance  $$\mathbb{Z}/6\mathbb{Z}=\{\overline{0},\overline{1},\dots,\overline{5}\}=\{\overline{0},\overline{1},\overline{1}+\overline{1},\dots, \overline{1}+\overline{1}+\overline{1}+\overline{1}+\overline{1}\}$$
$$\mathbb{Z}/6\mathbb{Z}=\{\overline{0},\overline{-1},\dots,\overline{-5}\}=\{\overline{0},\overline{5},\overline{5}+\overline{5},\dots, \overline{5}+\overline{5}+\overline{5}+\overline{5}+\overline{5}\}$$
In the case of $\mathbb{Z}$, this has only two generators, $1$ and $-1$, because other number, say $2$ for instance, this would only generate multiples of $2$, i.e. even numbers.
