How to prove that $\alpha^p = 1$ using induction?

Question

I am looking for an inductive proof of the following.

Let $\alpha \in S_n$ be a cycle of length $p$. Then $\alpha^p =1$.

For example $\alpha = (123)$ then $\alpha^3 = 1$ it is easy to verify.

My attempt: Induction on $p$: The base case is true for $p=1$. Inductive case, assume that true for $k$ i.e $\alpha^k = 1$. We want to prove for $\alpha^{k+1}$. Rewrite $\alpha^{k+1} $ can be rewritten as product of two cycle's of length less than $p$??

Note: I know that this question is already answered elsewhere on the site; however, I want to solve the above problem using induction.

Answer 1

Hint

I'm not sure that it's possible by induction. You can prove that all $p-$cycles are conjugate (it's a very common result), and thus, if $\alpha $ is a $p-$cycle, you can prove that $$|S_n/\left<\alpha \right>|=\frac{n!}{p}.$$

The claim follow.

Answer 2

I don't think that weak induction is a very good way of doing this, because elements of order $p$ and elements of order $p-1$ are not necessarily related in any way.

If you must use induction here, your best bet is to go for strong induction, and in this case, your base case will need to be that the property holds for all prime numbers $p$. There are a number of ways to prove this part, depending on what background you're allowed to assume.

After this, your inductive case will be to assume the property holds for all proper divisors of $p$, and show this implies the property for $p$. You can relate elements $x$ of order $p$ are to elements of order $k\mid p$ by noting that $x^{p/k}$ has order $k$.