I have the following question:
Let $G$ be a finite group. We define a relation $\sim$ on $G \backslash \left\{e\right\} = \left\{ g \in G : g \neq e \right\}$ by $g \sim h$ if and only if there exists $k$ such that $g = h^k$
1) Show that $\sim$ is reflexive and transitive
2) Show that $\sim$ is symmetric if and only if the order of $g$ is prime for every $g \neq e$.
1) Clearly $\sim$ is reflexive as for $k=1$ we have $g = g^1$ so $g \sim g$.
To show it is transitive, suppose $g \sim h$, so there exists a $k_1$ such that $g = h^{k_1}$, further suppose $h \sim l$ so there exists a $k_2$ such that $ h = l^{k_2}$ so we need to show that there exists a $k_3$ such that $g = l^{k_3}$ which implies $g \sim l$
I get a bit stuck at this point, because I'm unsure whether all $g \in G$ have finite orders or not. I know $G$ is a finite group, does this imply that all $g \in G$ are of finite order?
The working I've done so far, which assumes that all $g$ are of finite order, goes like this:
As $h$ is of finite order, let $o(h) = \lambda$
Since $g = h^{k_1}$ this implies that $g = h^{n\lambda + k_1}$ where $n \in \mathbb{Z}$
As $l$ is of finite order, let $o(l) = \mu$
Likewise we get that $h = l^{m\mu + k_2}$ where $m \in \mathbb{Z}$
So $h^{\lambda+1-k_1} g = h = l^{k_2}$
$g = h^{k_1 -1} l^{k_2}$
It's at this point I don't really know how to go forward (which is making me think I've tackled the problem incorrectly, but I don't see another way)
Thanks, any help appreciated!
For part (2), you can use the formula that says that if $\lvert x \rvert$ is the (finite) order of the element $x$ of the group $G$, then $$ \lvert x^k \rvert = \frac{\lvert x \rvert}{\gcd(k, \lvert x \rvert)}.\tag{ordpow} $$ So if the order of $x$ is not a prime, write $\lvert x \rvert = m n$, with $m, n > 1$. Then $1 \ne y = x^ m$ has order $n < m n$, and there's no way to write $x$ as a power of $y$.
On the other hand, if $x$ has prime order $p$, and $1 \ne x^k = y$, then $p$ does not divide $k$, so $\gcd(p,k) = 1$. Find $u,v$ such that $p u + k v = 1$, and then $$ x = x^1 = x^{p u + k v} = (x^k)^v = y^v. $$