I was going through the definition of "order of an element" and in the examples following it the author had considered the group $U(15)$ under multiplication modulo $15$, where it was calculating the order of each element there it was given, $7^1 = 7$, $7^2 = 4 $... Now I understand that $7^1 = 7\pmod{15} = 7$. But In group theory where we talk about binary operations what does $g^1$ even mean? Don't we need two elements to perform Binary operation.
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3$\begingroup$ If $g$ belongs to a group $G$, $g^1=g$. $\endgroup$– José Carlos SantosDec 4, 2021 at 11:14
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1$\begingroup$ But then $g^2$ represents $g*g$ where $g \in G$ so why represent it as $g^1$ plus it writes $7^1$ = following it up with $7^2$ = 4 wouldn't it imply that $7^1 = 7mod15 = 7$ $\endgroup$– greyDec 4, 2021 at 11:24
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3$\begingroup$ If $G$ is a group written multiplicatively with identity $e$, for any $g\in G$ we define recursively $g^0=e$, $g^{n+1}=g^ng$, and for $n\gt 0$, $g^{-n}=(g^n)^{-1}$. It's notation. If $G$ is written additively with identity $0$, then we define $0g=0$, $(n+1)g=ng+g$, and $(-n)g=-(ng)$. Again, this is just notation. $\endgroup$– Arturo MagidinDec 4, 2021 at 14:14
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1$\begingroup$ I don't understand the downvotes here. This is a good question - the notation $g^1$ is ubiquitous in group theory, yet as the question points out it makes no sense. $\endgroup$– user1729Dec 4, 2021 at 16:58
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2$\begingroup$ @user1729 I didn't downvote, but saying "it makes no sense" is rather an overstatement. If you have a multiplicative structure, there is an obvious meaning to something like "$g^2$", "$g^3$"; and we make "$g^{-1}$" have a meaning that is completely consistent with that, so there is no obstacle to extending this in an obvious manner to $g^1$, just like we do in basic algebra and calculus. There's nothing wrong with being confused, but to assert that the notation "makes no sense" is taking it too far. $\endgroup$– Arturo MagidinDec 4, 2021 at 22:52
1 Answer
Generally if $G$ is a group with the neutral element $e\in G$, $g\in G$ is any element and $n\in\mathbb{Z}$ is an integer then $g^n$ is defined by the following rules:
$$g^0:=e$$ $$g^n:=gg^{n-1}\text{ for }n>0$$ $$g^n:=(g^{-1})^{-n}\text{ when }n<0$$
The middle rule is recursive. In the last one note that when $n<0$ then $-n>0$ and so $(g^{-1})^{-n}$ is well defined by the previous rule.
Few examples that follow from this definition:
$$g^1=gg^0=ge=g$$ $$g^2=gg^1=gg$$ $$g^3=gg^2=ggg$$ $$g^{-2}=(g^{-1})^2=g^{-1}g^{-1}$$ $$g^{-3}=(g^{-1})^3=g^{-1}g^{-1}g^{-1}$$
and so on.
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1$\begingroup$ $g^1$ is not a binary operation and by definition in group theory there is an binary operation on the elements of the same set. $\endgroup$– greyDec 4, 2021 at 11:27
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2$\begingroup$ $g^1$ is defined using the binary operation $\cdot$ of the group as $g^1 = g\cdot g^0 = g\cdot e$. $\endgroup$ Dec 4, 2021 at 11:34
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$\begingroup$ @user1729 I suppose the discussion about "binary operation" term indeed is irrelevant. I've removed that part. $\endgroup$– freakishDec 4, 2021 at 15:35
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$\begingroup$ OK, great, I've removed my comment and added a +1 :-) $\endgroup$– user1729Dec 4, 2021 at 15:58