Question on order of elements in groups (subgroups)

Question

I am a bit confused at the moment, but what can we say about the order of all elements in a finite (sub)group? Suppose we have a group $G$ such that $|G|=p^k$ for a prime $p$. Next let $H$ be a subgroup then $|H| = 1$ or $p$ or $p^2$ or ... up to $p^k$. Assume the subgroup's order is $p^i$ for some $i<k$.

So here is my confusion: what can we say about the order of the elements in $H$? As far as I remember there is a theorem that states that $$o(a) \;|\; |H|$$ i.e. $o(a)$ must divide the (sub)group order. This would mean that in my example $o(a) = 1$ or $p$ or $p^2$ or ... $p^i$.

Now my questions:

1. $o(a)$ must divide $|H|$, but also divide $|G|$, would this imply that $o(a) = gcd(|H|,|G|)$ ?
2. Let $H=\{e,a,b\}$ be a group where $a^2 = b^2 = e$, then $o(a) = 2$ but 2 does not divide 3, what is wrong here?

Answer 1

1- Notice that $|H|$ divides $|G|$ so $\gcd(|H|,|G|)=|H|$.

2- There's not a group $H$ with three element such that $a^2=e$ since in this case $o(a)=2$ but $2$ doesn't divide $|H|=3$.

Answer 2

For the first question, consider a group of the form $\Bbb Z/p\Bbb Z\times \cdots\times\Bbb Z/p\Bbb Z$.

For the second question, you have elements $a$, $b$, $e$, and $ab$. We can see $ab$ is distinct from the other elements since otherwise, say, $ab=a$, and multiplication on the left by $a$ yields $b=e$. In fact, this group is isomorphic to $\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$.