Are all cyclic groups considered sub-groups of another group?

Question

Consider $Z_5^*$ as a group G, where the group operation is multiplication. Using 2 as a generator and element of G, a cyclic group can be generated:

• 2⋅0 (mod 5) = 0
• 2⋅1 (mod 5) = 2
• 2⋅2 (mod 5) = 4
• 2⋅3 (mod 5) = 1
• 2⋅4 (mod 5) = 3
• 2⋅5 (mod 5) = 0

Is this cyclic group and all cyclic groups that can be generated from $Z_5^*$ considered a sub-group of $Z_5^*$?

Answer 1

What you have written is not a cyclic group. Given that the operation in $Z^\ast_5$ is multiplication, what you want is repeated multiplication by 2, i.e. taking powers, which gives you $2^0 =1$, $2^1=2$, $2^2= 4$, $2^3=8\equiv 3$, $2^4=16\equiv 1$.

cyclic groups generated this way are considered subgroups of $Z^\ast_5$. In this instance, the subgroup is the whole group!

Answer 2

As @RyleeLyman pointed out, the list should be denoted as

$$ 2^1 = 2 \implies 2 \equiv_{5} 2\\ 2^2 = 4 \implies 4 \equiv_{5} 4\\ 2^3 = 8 \implies 8 \equiv_{5} 3\\ 2^4 = 16 \implies 16 \equiv_{5} 1.\\ $$

The reason being that you are considering the multiplicative group $\mathbb Z_{5}^*$.

Is this cyclic group and all cyclic groups that can be generated from $\mathbb Z_5^*$ considered a sub-group of $\mathbb Z_5^*$?

$\mathbb Z_5^*$ is a subgroup of $\mathbb Z_5^*$, since every group $G$ is a subgroup of itself. [1]

Another way of looking at this is to use the fact that all subgroups of a cyclic group are cyclic. [2] In other words, the cyclic group that can be formed under the multiplicative group of $\mathbb Z_5^*$, is $\mathbb Z_5^*$ itself using the generator $2$,

$$ \mathbb Z_5^* = \langle 2 \rangle. $$