Order of an element in a group problem as follows

Question

Let $(G,*)$ be a group, $a \in G$, and $|a|=p$, where $p$ is a prime.

1. Prove that $|a^k|=p$ for all $1 \le k \lt p$.
2. Prove that for all $m \in \Bbb N$, either $a^m = e_G$ or $|a^m|=p$, where $e_G$ is the identity element in $G$.

attempt:

1. Suppose $|a^k| \ne p$ for all $1 \le k \lt p$. Then, $a^{pk} \ne e_G$. On the other hand, $|a|=p$ i.e. $a^p = e_G$. So, we have \begin{equation*} a^{pk} = (a^p)^k \ne e_G \Leftrightarrow e_G^k \ne e_G \end{equation*} i.e. $e_G \ne e_G$, a contradiction. Hence, $|a^k| = p$.

Alternatively, by using theorem: Let $(G,*)$ be a group, $a \in G$, and $|a| = n$. Then, for all $t \in \Bbb N$, \begin{equation*} |a^t| = \frac{n}{\gcd(t,n)} \end{equation*} by replacing $t=k$ and $n=p$, we have: If $p \nmid k$, then we're done. If $p \mid k$, then $|a^k| = \frac{p}{p} = 1$, a contradiction. Hence, we must have $p \nmid k$ i.e. proved.

2. Note that for all $m \in \Bbb N$, \begin{equation*} \gcd(m,p) = \begin{cases} p, \text{if $p \mid m$} \\ 1, \text{if $p \nmid m$} \end{cases} \end{equation*} Then, by theorem above, we have \begin{equation*} |a^m| = \frac{p}{\gcd(m,p)} = \begin{cases} p, \text{if $p \mid m$} \\ 1, \text{if $p \nmid m$} \end{cases} \end{equation*} Thus, either $|a^m| = 1 \Rightarrow a^m = e_G$ or $|a^m| = p$, as desired.

Does the above proof true? I still have a little bit confuse in proving 1. Any elementary or a simple approach?

Answer 1

I think you might be having some misunderstanding about the statements. In your first proof for statement 1, you tried a proof by contradiction. To do so, you need to assume that the original statement is false. What does it mean for "$|a^k| = p$ for all $1 \leq k < p$" to be false? It means there exists a $k$ with $1 \leq k < p$ and $|a^k| \neq p$. It does not mean $|a^k| \neq p$ for all $k$. So your first proof is not correct.

In your second proof, you made a small mistake: it is not possible that $p|k$, because $k < p$. Thus, there is no "contradiction". You simply stop before your last sentence and the proof is concluded.

Your proof for the second statement is correct.