1

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 : )

Shaun
  • 44,997
Dhiraj Rao
  • 77
  • 3
    I think your understandings are correct. But one mustn't take the ordinary english meaning of "generated" too seriously: the mathematical meaning (with powers of $a^{-1}$ allowed) is what we use in mathematics. – ancient mathematician Oct 25 '22 at 11:16
  • 4
    A few points: 1) For finite groups you don't need inverses, so everything is fine there. 2) There are a few situations where we explicitly include all inverses of generators into the generating set, so this issue does come up, and it is solved that way. 3) 'Generated by' here means the smallest subgroup containing the generators is the whole thing. Since the subgroup always contains inverses, this is why we don't need to require them. Not including inverses is called the 'semigroup generated by', and also appears in some parts of group theory (infinite permutation groups). – David A. Craven Oct 25 '22 at 11:36
  • 4
    But I wouldn't call $a^{-n}$ an abuse of notation. For a positive integer $n$, we define $a^{-n} := (a^{-1})^n$. An abuse of notation is where you deliberately use notation that is technically incorrect, but easily understood by anyone familiar with the subject area, and designed to avoid excessively cumbersome (but formally correct) notation. – Derek Holt Oct 25 '22 at 11:40
  • Perhaps relevant: https://math.stackexchange.com/questions/4412683/proofs-request-proofs-that-five-exponention-rules-hold-for-positive-real-bases/4412691#4412691 – Ethan Bolker Oct 25 '22 at 11:49
  • 2
    I like to also think of $a^{-n}$ as shorthand for $(a^n)^{-1}$ which means the inverse of $a^n$. But whichever way you think of it, the meaning of $a^{-n}$ is completely unambiguous in the context of group theory, so it's not what you should call an "abuse of notation" which usually connotes ambiguity or actual contradiction. I would instead call it an "extension of notation", which is extremely common throughout mathematics, given the limited number of symbols and words we have to work with. – Lee Mosher Oct 25 '22 at 11:55
  • Yes you are correct. To make it definitive, I'd suggest to use the ":=" notation ("is a symbol for"): $a^{-n}:=(a^{-1})^n$. – citadel Oct 25 '22 at 13:19
  • Don't forget to prove that $$(a^n)^{-1}=(a^{-1})^n.$$ – Michael Hoppe Oct 25 '22 at 17:00

2 Answers2

1

The thing to keep in mind here is that the group operations of taking inverses commutes with rising to a power: $(a^n)^{-1} = (a^{-1})^n$. To see this you can multiply both sides of the equation by $a^n$. Therefore, we can call both operations $a^{-n}$.

Santi
  • 91
-2

remember that since $G=gp(a)$ is a group then $a^{-1}\in G$.

Thus: $a^{-1}...a^{-1}$ $n$ times is $a^{-n}$

ryaron
  • 1,091