# Is every power of a group element also contained in the group?

If we know that $a$ is in a group, then any power of $a$ would also be in the group because it is closed under the group operation, right?

• Yep, that's right (including the case of negative exponents) – Hagen von Eitzen Mar 22 '15 at 16:22
• You're correct. Moreover, if the group is of finite order, there must exist $i$ and $j$ such that, at some point, $a^i = a^j$. More info here – Miguelgondu Mar 22 '15 at 16:30
• Where else would it be? – Slade Mar 22 '15 at 22:49
• Yes, for any integer power. – paw88789 Mar 22 '15 at 22:59

A binary operation on a set $S$ is a function $f$ with domain $S\times S$ and co-domain $S$. This means any combination of the group's operation stays in the group.
If you want a formal proof, $a=a^1$ is in the group by your hypothesis, and assuming that $a^{i-1}$ is in the group it follows that $a^i = a a^{i-1}$ is in the group. So by induction, $a^n$ is in the group for all $n$.