Let $n\in\mathbb{N}$ and $a$ be an element of a group $(G,*)$. Then, as far as I understand, $a^n$ is a shorthand notation for $a*...*a$ ($n$ times).
Similarly, $(a^{-1})^n$ is a shorthand notation for $(a^{-1})*...*(a^{-1})$ ($n$ times).
It gets a bit tricky for me when I see $a^{-n}$ (again $n\in\mathbb{N}$)
Because $a*...*a$ ($-n$ times) makes no sense, I take it that $a^{-n}$ is the same as saying $(a^{-1})*...*(a^{-1})$ ($n$ times). This is consistent with $a^{-n}=(a^{-1})^n$.
Please let me know whether my understanding is right or wrong.
My motivation behind seeking the above clarification:
Let $G=\langle a \rangle$ be an infinite cyclic group generated by $a$. Then by definition, any element $g$ in $G$ can be written as $a^m$ for some $m\in\mathbb{Z}$. Then one of the following is true:
- $g=a*...*a$ ($m$ times) if $m$ is positive
- $g=(a^{-1})*...*(a^{-1})$ ($|m|$ times) if $m$ is negative
Let us ignore the case where $m=0$ as there is no confusion there.
Now from my classification above, can I say that any element of $G$ is a product of finite number of $a$'s OR a product of finite number of $a^{-1}$s. What I find awkward is that even in the latter case, we say the element is generated by $a$ although in some sense that element is generated by $a^{-1}$.
Someone please help me discard my awkwardness : )
remember that since $G=gp(a)$ is a group then $a^{-1}\in G$.
Thus: $a^{-1}...a^{-1}$ $n$ times is $a^{-n}$