TITLE: Show that $x$ has larger order than $a$
QUESTION [5 upvotes]: This is exercise $2.35$ from Rotman's A First Course in Abstract Algebra.
Let $G$ be a group and let a $a \in G$ have order $pk$ for some prime $p$, where $k \geq  1$.
Prove that if there is $x \in G$ with $x^p = a$, then the order of $x$ is $p^2k$, and hence $x$
has larger order than $a$.
This isn't homework, but I'm stuck. Is there a nice way to prove it?

REPLY [2 votes]: This the special case $n = 1$ of the following fact:

Let $p$ be a prime, and suppose $a \in G$ has order divisible by $p$ and $x^{p^n} = a$. Then $x$ has order $p^n \cdot o(a)$.

To prove this, you can use the order formula $$o(x^{p^n}) = \frac{o(x)}{\gcd(o(x), p^n)} $$
Thus $o(x) = \gcd(o(x), p^n) \cdot o(a)$. Let $p^k$ be the largest power of $p$ dividing $o(x)$. If $k < n$, then $o(x) = p^k \cdot o(a)$. Because $p$ divides $o(a)$, this implies that $p^{k+1}$ divides $o(x)$, a contradiction. Hence $k \geq n$, which proves that $o(x) = p^n \cdot o(a)$.