Let $G=\left \langle x \right \rangle$ a cyclic group s.t. $ord(x)=n$. Let $a=x^\alpha$ s.t. $\alpha \nmid n$ and $0<\alpha<n$ . Prove that $x^\alpha$ is a generator of $G$.
My partial answer: Let $A:=\left \langle x^\alpha \right \rangle$. We'll assume in counterwise the oposite. Then, $|G|=n$ thus $|A|\leq n$. $|A|\ne n$ because elsewise we would get $\left \langle x^\alpha \right \rangle = G$. Thus, $|A|<n$. I would like for some help from here or to get another direction.
On
This is not true. For example, suppose $\text{ord}(x)=10$, and take $\alpha=6$. Then $$<x^\alpha>=\{e, x^6, x^{12}=x^2, x^8, x^{14}=x^4\}$$ which clearly is not the whole group. However, if $\gcd(\alpha,n)=1$ this is true, since then by the euclidean algorithm there exist $k,l \in \mathbb{Z}$ such that $k\alpha+ln=1$, so $$x=x^{k\alpha+ln}=x^{k\alpha}x^{ln}=(x^{\alpha})^k (x^n)^l=(x^\alpha)^k$$ where the last equality follows from lagrange’s theorem (the order of $x$ divides the order of the group). The above equality now shows $x$ is in the subgroup generated by $x^{\alpha}$, so the subgroup contains all of $G$.
On
That is false.
Counter-example:
The multiplicative group $\;\bigl(\mathbf Z/11\mathbf Z\bigr)^{\times}$ is cyclic of order $10$, generated by the congruence class of $2$. However, $x=2^4$ has order $5$ (a divisor of $10$): modulo $11$, $$a=5,\enspace a^2=3,\enspace a^3=4,\enspace a^4=9,\enspace a^5=1.$$
The correct assertion is thatit is true if and only if $\gcd\alpha,n)=1$. More generally, if $x$ generates a cyclic group of order $n$, $a=x^\alpha$ generates a cyclic group of order $$\frac n{\gcd(\alpha,n)}$$
You can't prove it, since it is not true. Take $(\mathbb{Z}_6,+)$, which is a cyclic group of order $6$ and generated by $1$. Take take $\alpha=a=4$. Notice that $4<6$ and that $4\nmid6$. However, $\langle a\rangle=\{0,2,4\}\neq\mathbb{Z}_6$.