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- Cyclic Group Generators of Order $n$ 5 answers
Let us say that $n = 5$. The number of generators of a finite cyclic group would be the number of numbers that are relatively prime to $n$ and the identity element.
Here, when $n=5$, the number of generators would be $2,3,5$. We would also have to include the identity element generator, $e$ for each finite cyclic group. Thus the number of generators of a cyclic group with order $5$ would be 4.
Similarly, when $n=12$, the number of relatively prime generators would be $2,3,5,7,11$ and $1$ for a total of 6.
I think I understand how to find the number of generators of a cyclic finite group, but I can't really explain why.
I also can't draw the connection to this statement:
The order of an element in a finite cyclic group is the smallest positive integer $n$ such that $a^n=e$, denoted ord $a$.
How would I prove that the order of an element $a \in G$ equals the order of $G$? What am I failing to consider?