# Proving that groups are cyclic and finding generators.

I would just like to ask for a quick explanation on how one actually proves a group is cyclic and how to find the generator functions.

So the questions says:

Let $$G_1$$ be a group defined on the set $$\mathbb Z_4$$ with binary relation $$+$$ (addition modulo $$4$$), and let $$G_2$$ be a group defined on the set $$\mathbb Z_5 − {[0]}$$ with binary relation $$\cdot$$ (multiplication modulo $$5$$).

Prove that both $$G_1$$ and $$G_2$$ are cyclic and write down generators for each group.

I started by writing out the respective modulo tables for each group.

So for $$(\mathbb Z_4 , +)$$ you have the elements $$\{0,1,2,3\}$$ and for $$(\mathbb Z_5 - {[0]}, \cdot)$$ you have the elements $$\{1,2,3,4\}$$.

As far as I am aware, a cyclic subgroup of a group $$(G , \star)$$ is a set $$\langle a \rangle = \{a^k : k \in \mathbb Z\}$$, which is equal to $$\{\dots, a^{-2}, a^{-1}, a^0, a^1, a^2, \dots \}$$ and we call this the set generated by $$a$$.

I have the tables and definitions, but where do I go from here?

• Just pick any element $a$ as a potential generator. Then calculate $a*a$, $\ \ a*a*a$, $\ \ a*a*a*a$, ... where $*$ is the group operation, and see what happens. Do you get in this way all the elements of the group from $a$? What if I pick a different element $a$, will that one turn out to be a generator? May 17 at 8:56

As a general rule, you prove a group is cyclic by exhibiting a generator. So you find an element $$a\in G$$ such that $$G=\{a^n; n\in \Bbb{Z}\}$$ or $$G=\{na;n\in\Bbb{Z}\}$$ when the group is noted additively.
So for $$(\Bbb{Z}_4,+)$$, $$1$$ is definitely a generator because
$$\{0,1,2,3\}=\{1,1+1,1+1+1,1+1+1+1\}$$
and for $$(\Bbb{Z}_5,\cdot)$$, $$2$$ is definitely a generator because
$$\{1,2,3,4\}=\{2,2^2,2^3,2^4\}$$
• Yes but they are the same so $1$ and $3$ generate the same subgroup, in fact the whole group while $0$ generates the trivial subgroup and $2$ generates the proper subgroup $H=\{0,2\}$ May 22 at 9:24