# Meaning of permutation representation

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 $$b\in S$$?

• 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) – J. W. Tanner Feb 6 at 3:47
• @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. – Tobias Kildetoft Feb 6 at 8:02
• Please ask one question at a time. – Shaun Feb 6 at 9:37

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 $$\{a,b\}$$.
• 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. – Ted Feb 7 at 6:38