# Generators of a finite cyclic group.

I have to find a single generator in the form G = for the finite cyclic group G = <13, 20>. I'm having trouble figuring out what the group <13,20> means and how to simplify this. I found this - (Generators of a cyclic group) as a related concept but am having trouble interpreting the results in the lemma there in the question I am given.

G is a subgroup of < Z, + >

Another clue I've figured out is the following line from the wikipedia page for Generating Set of a Group - "Different subsets of the same group can be generating subsets; for example, if p and q are integers with gcd(p, q) = 1, then {p, q} also generates the group of integers under addition (by Bézout's identity)." --- I'm again not sure how to interpret this in finding what the group itself is.

I'd appreciate clues to get me a step or two further, thanks!

• I don't understand what you mean by $\langle 13,20\rangle$. This doesn't inherently mean anything. Do you mean generated by an element of order $13$ and an element of order $20$? – Ian Coley Jan 30 '14 at 23:50
• What is the group operation? – David Jan 30 '14 at 23:51
• Well, what I mean by <13, 20> (as it is given in the book I am studying from) is that it is a cyclic subgroup of <Z, +> finitely generated by the set {13, 20}. I have not been able to get an understanding of what the meaning of the subgroup generated by a set of more than one element is, which is the root cause of my problem. – user125080 Jan 30 '14 at 23:57
• Sorry! As I put in my last comment, group operation is addition. I'll edit that into the question. – user125080 Jan 30 '14 at 23:57

Yes, Bezout's identity is useful here.

By the comments, it seems that you are looking for the finite-ly generated subgroup $\langle 13,20\rangle$ of the group $(\Bbb Z,+,-)$ which is also claimed to be cyclic.

So, if $13,20\in H$ for any subgroup $H$ of $\Bbb Z$, then also $7=20-13$ and $6=13-7$ and hence $1=7-6$ are all in $H$ as $H$ is closed under subtraction. But if $1\in H$ then every $n\in\Bbb Z$ is in $H$ (as $n=\pm(1+1+1+\dots)$), so $H=\langle 1\rangle=\Bbb Z$.

In particular, the generated subgroup $\langle 13,20\rangle$ can also be generated by $1$.

In general, by Bezout's identity, we have $$\langle a,b\rangle = \langle\gcd(a,b)\rangle$$ in $(\Bbb Z,+,-)$.

I think $\langle 13, 20 \rangle$ is the subgroup of $(\mathbb{Z},+)$ generated by $13$ and $20$. I.e., $$\langle 13, 20 \rangle=\{13a+20b:a,b \in \mathbb{Z}\}$$ under addition.

In this case, $\gcd(13,20)=1$, so using the claim mentioned in the question, $$\langle 13, 20 \rangle=\mathbb{Z}.$$

So, what's a generator of $\mathbb{Z}$?

Let $a\in \mathbb{Z}$. Note that $\mathbb{Z}$ is an additive group. The the subgroup of $\mathbb{Z}$ generated by $a$ is $$\langle a\rangle = a\mathbb{Z} = \{\dots,-3a, -2a, -a, 0 , a, 2a , 3a, \dots\}.$$ The group $H = \langle a,b\rangle$ is a group containing $a$ and $b$. This means that $H$ contains all numbers of the form $na + mb$ for any $n,m\in \mathbb{Z}$. This means that if you can find $n$ and $m$ such that $n13 + m20 = 1$, then $1\in \langle 13,20\rangle$. And if the group contains $1$, then ...