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?

  • $\begingroup$ 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? $\endgroup$ May 17 at 8:56

1 Answer 1


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


and for $(\Bbb{Z}_5,\cdot)$, $2$ is definitely a generator because


  • $\begingroup$ Hi what about Cyclic subgroups? There are two generators for the modulo 4 example right, which are {1,3}. Is the cyclic subgroup just the element generated by the generator? for example, the cyclic subgroups for G1 are {0, 1, 2, 3} (generated by 1) and {0, 1, 2, 3} (generated by 3). Am i right? $\endgroup$ May 21 at 12:22
  • $\begingroup$ 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 $H=\{0,2\}$ $\endgroup$
    – marwalix
    May 22 at 9:24

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