# Cyclic groups and other groups

Ok So I know that a cyclic group is a group that is generated by a single element like $\large{(Z_n,+)}$.

Now I was wondering that if every group has a generator , and I found that the answer is yes in here.

However, This answer doesn't make sense. He is saying that every group has a set of generators and a group that has only one generator is cyclic.

Does that mean $\large{(Z_7,+)}$ is not cyclic for instance because both $1$ and $3$ are both generators of $\large{Z_7}$. But here in this website, they state that $\large{(Z_7,+)}$ is a cyclic group.

So in Summary here is my questions

(Question 1)

Is it true that if a group has more than one generator then it is not cyclic ?

(Question 2)

I want to see an example where a group has a set of generators not necessarily a single element !

I think that it is ok to have a cyclic group that is generated by more than one element as long as they are single elements, but how can we have a generator which is not a single element, I am really confused.

Question $1$:

No, that is not true. The statement you are looking for is that if a group cannot be generated by a single element then it is not cyclic. In other words, if there does not exist a $g \in G$ such that $G = \{ g^n \ | \ n \in \mathbb{N} \}$, then $G$ is not cyclic. Contrapositively, $G$ is cyclic if we can find a $g \in G$ such that $G = \{g^n \ | \ n \in \mathbb{N} \}$. But the wording here is subtle: notice in particular that we don't specify that $g$ be unique! So $\mathbb{Z}_7$ is indeed cyclic even though both $1$ and $3$ generate it.

Question 2:

• The example that immediately comes to mind is the dihedral group of order $2n$, which is the group of symmetries of a regular $n$-gon. This group is not cyclic, but it can be generated by $2$ elements: $R$ and $F$, where $R$ is a rotation of $\displaystyle \frac{2\pi}{n}$ radians about the center, and $F$ is any reflection over a line of symmetry. To keep things simple, confirm this for a triangle or a square, and then try generalizing.

• Another example is the Klein-$4$ group $\mathbb{Z}_2 \times \mathbb{Z}_2$. This group is not cyclic, but it can be generated by two elements, for example, $(0, 1)$ and $(1, 0)$ as you can check.

• A rather complicated example is that of $S_p$ for some prime $p$. This is the group of permutations on $p$ elements. This group can be generated by any $2$-cycle together with any $p$-cycle.

• Ok for the Klein-4 group, How can $(0,1)$ and $(1,0)$ generate the group ? we have $(0,0),(0,1),(1,0),(1,1)$ in the group. Now if I multiply $(0,1)(1,0)$ I get $(0,0)$. But how do I get $(1,1)$ ? – alkabary May 15 '15 at 5:42
• Careful! The group operation is addition modulo $2$, so $(0, 1) + (1, 0) = (1, 1)$. – Kaj Hansen May 15 '15 at 5:44
• And of course we already have $(0, 1)$ and $(1, 0)$. To get $(0, 0)$, we simply take $(0, 1) + (0, 1) = (0, 2) = (0, 0)$. – Kaj Hansen May 15 '15 at 5:46
• hmmm, ok so $(0,1) + (1,0) = (1,1)$ and $(0,1) + (1,0) + (0,1) + (1,0) = (1,1) + (1,1) = (0,0)$ hmm that make sense ! Thanks a ton! – alkabary May 15 '15 at 5:46
• I'm glad I could help! – Kaj Hansen May 15 '15 at 5:48

Question 1: A group $G$ is cyclic if and only if $G$ can be generated by one element. This element does not have to be unique. So $(\mathbb{Z}_7,+)$ is indeed cyclic.

Question 2: ($\mathbb{Z}_2 \times \mathbb{Z}_2,+)$ is not cylic. It is of order 4, yet no element has more than order 2.

• What are the set of generators of $Z_2 \times Z_2$ ? – alkabary May 15 '15 at 5:37
• Three Possibilities: $\{(1,0),(0,1)\}$ or $\{(1,0),(1,1)\}$ or $\{(0,1),(1,1)\}.$ – matt biesecker May 15 '15 at 5:39
• A good exercise for you to work out on your own is to prove any cyclic group is Abelian. Then take it a step further and show that any cyclic group is isomorphic to $\{\mathbb{Z}_n,+)$ for some $n$ or is isomorphic to $(\mathbb{Z},+).$ – matt biesecker May 15 '15 at 5:43
• @mattbiesecker Those are the minimal generating sets. You could also take any subset containing any of those. – Derek Holt May 15 '15 at 7:54
• @DerekHolt. Thanks. Getting sloppy in my old age. – matt biesecker May 15 '15 at 8:12