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I'm following the discussion in Artin's Algebra. Artin defines a permutation representation of a group $G$ as a homomorphism from the group to the symmetric group:$$\varphi:G\rightarrow S_n$$

Then there's a proposition: There is a bijective correspondence between operations of $G$ on the set $S = \{1,..., n\}$ and permutation representations $\varphi:G\rightarrow S_n$ $$\Bigg[\textrm{ operations of G on S }\Bigg]\longleftrightarrow \Bigg[\textrm{ permutation representations }\Bigg] $$

Proof: Define a permutation representation $\varphi$ by setting $\varphi(g) = m_g$, left multiplication by $g$. Conversely, the same formula defines an operation of $G$ on $S$.

Question 1: Why does he only consider the operation "multiplication by $g$"? There are other operations like conjugation by $g$. Shouldn't these other operations also have corresponding perm. reps.?

Question 2: How about the operations of $S_3$ on the set of 2 elements $S = \{a, b\}$? If we are defining each homomorphism $\varphi(g) = m_g$ by left multiplying the set $S$ by $g \in S_3$, how does e.g. $(123) \in S_3$ act on $b\in S $?

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    $\begingroup$ left multiplication by $g$ permutes the elements of $G$. there could be a non-identity element that commutes with every element of $G$ and then conjugation would not be a permutation (a bijective, invertible map) $\endgroup$ – J. W. Tanner Feb 6 at 3:47
  • $\begingroup$ @J.W.Tanner That is not correct. Conjugation also gives a bijective map (it just has fixed points). But as explained in the answer, this is not really the issue here. $\endgroup$ – Tobias Kildetoft Feb 6 at 8:02
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    $\begingroup$ Please ask one question at a time. $\endgroup$ – Shaun Feb 6 at 9:37
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I think you have some things confused here.

The left side is the set of operations (also called actions) of $G$ on $S$, which are maps $G \times S \to S$ satisfying the properties of a group action. Given a $g \in G$, "left multiplication by $g$" means the map $s \mapsto gs$. Calling it "left multiplication" can be a bit misleading because there is no corresponding "right multiplication" (nor "conjugation" as you talk about in Q1), and it's not a "multiplication" in the sense of the multiplication in $G$. Remember that $G$ is acting on $S$, not on itself.

In Q2, your question is phrased a bit strangely because it sounds like you may be thinking there's only one way for $S_3$ to act on $\{a,b\}$. But that's not the case. $S_3$ can act on $\{a,b\}$ in any way that satisfies the properties of a group action. And $m_g$ is not defined until you have chosen an action of $G$ on $S$. Once you have chosen an action, then $m_g$ is the map sending $s \mapsto gs$; but $gs$ doesn't make sense unless you have chosen an action. In fact, there are two ways for $S_3$ to act on $\{a,b\}$.

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  • $\begingroup$ thank you sir. I was confused as to the meaning of the"multiplication" . So is the wording in artin ambiguous? He keeps referring to the operation as multiplication. $\endgroup$ – user35687 Feb 6 at 18:57
  • $\begingroup$ It's not ambiguous, as long as you understand that you are "multiplying" an element of $G$ with an element of $S$. I wouldn't normally call that multiplication though. $\endgroup$ – Ted Feb 7 at 6:38

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