List all elements having order $7$.

Question

Let $G= \langle a \rangle$ be a cyclic group of order $35$. List all elements having order $7$.

I am unsure how to find these elements. I know that they will all be of the form $a^m$, but how can I determine the values of $m$ that will satisfy the question?

Answer 1

The elements of $G$ can be written uniquely as $a^k$ for $0\leq k<35$. In order for $a^k$ to have order $7$, we need $a^{7k}=a^0$, so $7k$ has to reduce to $0$ modulo $35$. In other words, $k$ is a multiple of $5$. This gives $\{a^0,a^5,a^{10},a^{15},a^{20},a^{25},a^{30}\}$. However, the element $a^0$ has order 1 because it is the identity. Thus the elements of order $7$ are: $$ \{a^5,a^{10},a^{15},a^{20},a^{25},a^{30}\}. $$

Answer 2

By the Fundamental Theorem of Cyclic Groups, the only distinct subgroup of order d of a cyclic group <(a)> of order n, is <(a^(n/d))>, thus in this question, the only distinct subgroup of G is <(a^5)>. All subgroups of order 7 of G are <(a^(5*k))>, where 1<=k<=7, and g.c.d(k,7) =1. Thus, the subgroups are <(a^5)>, <(a^10)>, <(a^15)>, <(a^20)>, <(a^25)>, and <(a^30)>.