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For subset $S$ and $T$ of a group, define $ST = \{st|s \in S, t \in T\}$ and $S^T = \{s^t|s \in S, t \in T\}$.

What does $s^t$ mean in this context?

2 Answers 2

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The notion $s^t$ where $s,t$ are elements of a group denotes the conjugation, and is, as @BabakSorouh mentioned, equal to $t^{-1}st$

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    ...or $tst^{-1}$ depending on convention.2012-11-26
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    I think $s^t=t^{-1}st$ is the "better" convention: It makes $s^{tu}=(tu)^{-1}stu=u^{-1}t^{-1}stu=u^{-1}s^tu=(s^t)^u$. Compare with $s^{tu}=tus(tu)^{-1}=tusu^{-1}t^{-1}=(s^u)^t$, which is less attractive.2012-11-26
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    @HagenvonEitzen: I agree. But some people like superscripts to be contravariant and subscripts to be covariant. I wonder if there's any precedent for writing $s^t=tst^{-1}$ and $s_t=t^{-1}st$...2012-11-26
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    @CliveNewstead I thought I had rather seen $^ts$.2012-11-26
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    I have never seen $^ts$. I prefer $t^{-1}st$, but I wouldn't say it is more or less prevalent than $tst^{-1}$. It perhaps depends on the area you work in. Recently, writing $tst^{-1}$ makes what I have been doing much, much neater.2012-11-27
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    I have seen $s^t=t^{-1}st$ and $^ts=tst^{-1}$ in a paper that makes frequent use of conjugation.2013-04-16
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This symbol denotes conjugation: $s^t=t^{-1}st$ or $tst^{-1}$.

Be careful when you use this that you use a compatible notation with the commutator symbol. $s^t=t^{-1}st$ should be paired with $[s,t]=s^{-1}t^{-1}st$ and $s^t=tst^{-1}$ with $[s,t]=sts^{-1}t^{-1}$. The former is much more common, though the latter is seen sometimes too, but you don't want to mix them up or people will be real confused.

You'll also see $S^t$ going around, which means $S^t=\{s^t:s\in S\}$.

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    @CliveNewstead Haha, yes, sorry about that.2012-11-26