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Say I have elements $g$ and $h$ in a group $G$.

What does $g^h$ mean? Seeing this notation a lot but I can't find an explanation for it anywhere.

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@Jeremy:he or she say $h,g\in G$ – Maisam Hedyelloo Jun 12 '13 at 4:17
@DonAntonio Yes I realize that now, thanks to Maisam's comment – Jeremy Jun 12 '13 at 4:41
up vote 17 down vote accepted

In work by group theorists, this is the right action of $G$ on itself by conjugation: $$g^h = h^{-1} g h$$ This has the nice property that $$(gh)^k = g^k h^k \quad \text{and} \quad g^{(hk)} = (g^h)^k$$ The commutator associated with this is $[g,h] = g^{-1} g^h$, the difference between ${}^h$ and ${}^1$, the identity.

You will occasionally see other people use $g^h$ to mean $h gh^{-1}$ as a left-action. Sometimes this is called the topologist's convention, though we have some hope they will all adopt ${}^h g = h g h^{-1}$ so that their left action is on the left.

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I believe the idea to put it on the left is to get $^{hk}g = \hspace{-15 pt} \phantom{(g^k)}^h(^kg)$? – Patrick Da Silva Jun 12 '13 at 4:23
@PatrickDaSilva: exactly. Personally I use left actions when the group being acted on is additive, so $g(h+k) = gh + gk$ and $(gh)k=g(hk)$, and right actions when the group being acted on is multiplicative (as in my answer). – Jack Schmidt Jun 12 '13 at 4:25
I guess this other question really is relevant:… – Jack Schmidt Jun 12 '13 at 4:28

Usually, it means $h^{-1}gh$. That is, the application of the automorphism $\phi_h:G\to G$ which takes $g\to h^{-1}gh$.

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Let $X,Y$ be $G$-sets. Given a space $Y^X$ of maps $f: X \to Y$, the actions on $X$ and $Y$ induce an action on $Y^X$ given by $$g \cdot f(x) = gf(g^{-1}x)$$ where the $g\cdot f$ on the left hand side indicates the action on $Y^X$ and the $gf$ on the right indicates the action on $Y$. Obviously this notation is very ambiguous, so commonly we write $^g f$ for the action on $Y^X$. The symmetrical right action is generally denoted by an exponent on the opposite side.

Viewing a group element as a function $G \to G$ gives the etymology, so to speak, of this notation.

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The question is not about functions, as it happens. – anon Jun 12 '13 at 4:32
This doesn't have to do anything with the question. Please read it carefully and either edit or delete your post, @cameron – DonAntonio Jun 12 '13 at 4:33
Ah I was just trying to clarify the etymology, thanks for catching that. – Cameron Jun 12 '13 at 4:34

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