Question

If G = < x > is of order 25, then x^a generates G , where α is a factor of 25.
State the given statement is true or false, give reasons for your answer.

Answer 1

Answer on Question #78777 - Math - Abstract Algebra

July 2, 2018

Question. If $G=\langle x\rangle$ is of order 25, then $x^{a}$ generates $G$, where $a$ is a factor of 25. State whether the given statement is true or false, give reasons for your answer.

Solution. If we interpret the statement as “for every integer $a$ that is a factor of 25, $x^{a}$ generates $G$,” then the statement is false.

Let $a=5$ which is a factor of 25. We will prove that some element of $G$ is not in $\langle x^{a}\rangle$. Every element of $\langle x^{a}\rangle$ is of the form $(x^{a})^{n}$ for some integer $n$. We can divide $n$ by 5, so there are integers $q$ and $r$ such that $n=5q+r$ and $0\leq r<5$. We have

$(x^{a})^{n}=x^{an}=x^{5(5q+r)}=(x^{q})^{25}x^{5r}=x^{5r};$

$(x^{q})^{25}=e$ because the order of every element of $G$ divides 25. Hence $\langle x^{a}\rangle=\{x^{0},x^{5},x^{10},x^{15},x^{20}\}$, and this set contains at most 5 elements. However, $G$ contains 25 elements. Hence some element of $G$ is not in $\langle x^{a}\rangle$.

Answer. The statement is false.