# Generators of group $\mathbb{Z}_{4}$

Hello and sorry in advance for any mistakes, English isn't my first language. I recently started studying group theory for my university and I got introduced to cyclic groups. As an example, my book provides the group $\mathbb{Z}_{4}$ and says that numbers $1$ and $3$ are it's generators. Now please correct me if I am wrong, but wouldn't number $1$ being generator of $\mathbb{Z}_{4} \}$ mean that $\{1^n \mid n \in \mathbb{Z} \} = \mathbb{Z}_{4}$ ??

I can't understand why $1$ is generator of $\mathbb{Z}_{4}$. I'm assuming what i wrote above is correct, if not please correct me.

• Hint: What is the operation in $\mathbb Z_4$? What is $1^n$ in terms of that operation? Jun 30, 2017 at 21:28
• The operation is "+"! That you should keep in mind.
– Riju
Jun 30, 2017 at 21:28
• Oh god you are right..i was confused because it always takes the multiplicative notation for every operation. Hmm,$\mathbb{Z}_{4}$ isn't working for other operations? Jun 30, 2017 at 21:30
• It is not working for multiplication; in the sense that the structure will not be a group. You could in theory consider different compositions laws on $0,1,2,3$ so that you would get a group, but this is somewhat aritifically and in any case those laws would need to be specified;
– quid
Jun 30, 2017 at 21:35

When we say the group $\mathbb{Z}_4$, we're actually talking about $(\mathbb{Z}_4, +)$, meaning the operation over the group is $+$, not $\times$. Thus, 1 is a generator, because every element of $\mathbb{Z}_4$ can be written as $n \times 1= 1+1+1 + \dots$

Now you may ask, why wouldnt we give $\mathbb{Z}_4$ the $\times$ operation ? Well, what would be the inverse if 0 ? Of 2 ?

• You are so right..the book mentioned that it is going to use multiplicative notation for every operation. But cant $\mathbb{Z}_{4}$ exist with other operations? Jun 30, 2017 at 21:33
• Well, for instance, if you try to use the multiplication operation on $\mathbb{Z}_4$, you'll find that 0 and 2 have no inverses. (On the other hand, for any $n$, you can define a group structure on $\mathbb{Z}_n^*$ which is the set of elements of $\mathbb{Z}_n$ which have a multiplicative inverse.) Jun 30, 2017 at 21:35
• I'm sorry but aren't the elements of $\mathbb{Z}_{4}$ the residues of the division of an integer and $4$? So if we take $0$ as the identity element then wouldn't $2 \times 2 = 4mod4$ which is $0$ ? Wouldn't that make 2 the inverse of 2? I don't know if I am making any sense. Jun 30, 2017 at 21:44
• @Thomas The identity of the multiplication operation on $\mathbb{Z}_4$ would have to be 1. Jun 30, 2017 at 21:50
• Oh my god you are so right..I suck sorry I'm really new to this stuff i haven't gotten the hang of it yet.. thank you so much for helping Jun 30, 2017 at 21:54

The number $1$ generates $\mathbb{Z}_4$ because $\mathbb{Z}_4=\{1,1+1,1+1+1,1+1+1+1\}$.

Formally speaking, we define a group to be an ordered pair of a set $G$ coupled with a binary operation $*$ and satisfies the group axioms, which we write as $(G,*)$. In this case, our set is $\mathbb{Z}_4$ (set of residues modulo $4$) and our operation is $+$, so we can write the group as $(\mathbb{Z}_4,+)$. It is imperative to specify the binary operation on the set, as it is intrinsic to the definition of group.

In general, it is easy to see that for any additive group $\mathbb{Z}_n$ with $n \in \mathbb{Z}_{\geq 2}$, $1$ must be a generator as we can execute the addition operation $1, 2, ..., n-1$ times, where at the $(n-1)$-th iteration, we reduce modulo $n$ to get $0$.

1 and 3 ar all the generators of $$\mathbb{z}_4$$:

$$<1> = \{ 1, 2, 3 , 4 \equiv 0 \}$$ is a cyclic group, and also $$<-1>=<3>$$

$$<3> = \{ 3 , 6 \equiv 2, 5 \equiv 1, 4 \equiv 0 \}$$

Also you can do this:

we want a number $$n \in \mathbb{Z}_4$$ such that $$gcd(n , 4) = 1$$. So n=1 or n = 3 (generators).

(srry my english, I'm learning because I speak Spanish)