0

Disclaimer: my question is different from this question.

I am following these math notes on abstract algebra that I found online to teach myself. On page 21 the group action of conjugation is defined as: $$\mu:(x,y)\mapsto x^{-1}*y*x$$ In the other question it is defined as: $$\mu_\text{(other question)}:(x,y)\mapsto x*y*x^{-1}$$ Which works out.

Since we require: $$\mu(x*y,z)=\mu(x,\mu(y,z))$$ I find that the LHS is: $$\mu(x*y,z)=(x*y)^{-1}*z*(x*y)=y^{-1}*x^{-1}*z*x*y$$ Whereas the RHS is: $$\mu(x,\mu(y,z))=\mu(x,y^{-1}*z*y)=x^{-1}*y^{-1}*z*y*x$$ So clearly $$\mu(x*y,z)\neq\mu(x,\mu(y,z))$$

Is there a mistake on my end or did the author mean to define it as: $\mu:(x,y)\mapsto x*y*x^{-1}$?

Shaun
  • 44,997
QPhysl
  • 102
  • 5
    A good mental model for conjugation is changing coordinates (this can be made precise). Your question comes down to this: should we think of a rotation as spinning the paper or moving the points? You'll sometimes hear these talked about as passive vs. active transformations. Can you work out which is which in this case? – Charles Hudgins Aug 11 '23 at 23:37
  • 1
    @CharlesHudgins In terms of transforms, I think of the inverse occurring on the right. Because our new coordinate needs to be brought back to the original frame, acted on, and then transformed into the new frame. This seems to suggest there is a typo and it should be $\mu:(x,y)\mapsto xyx^{-1}$. Unless the typo is in the condition $\mu(x*y,z)=\mu(x,\mu(y,z))$ which needs to be changed? – QPhysl Aug 11 '23 at 23:44
  • 8
    It depends on whether you mean a left group action or a right group action, I think. Does the question at https://math.stackexchange.com/questions/2261660/is-there-a-standard-notation-for-conjugation help? – Steve Kass Aug 11 '23 at 23:47
  • 2
    @QPhysl I realize I misread your question. I thought you were wondering about the distinction between $xyx^{-1}$ and $x^{-1}yx$. Yes, Steve Kass is right. The difference in definition comes down to left vs. right group actions. – Charles Hudgins Aug 11 '23 at 23:51
  • You defined $\mu(x,y)$ to have the inverse on the right, but then when you evaluate $\mu(x*y,z)$ you put the inverse on the left... – coiso Aug 11 '23 at 23:58
  • @coiso, No I specifically define it on the left and continue this every time except when I show $\mu_\text{(other question)}$. It seems most people put it on the right which is why I brought that point up – QPhysl Aug 12 '23 at 00:02
  • @CharlesHudgins, sorry I don't exactly follow the difference. The resource I'm using defines the group action as satisfying two axioms, the first is: $\forall x,y\in G, s\in S,\mu(x*y,s)=\mu(x,\mu(y,s))$ for $\mu:G\times S\rightarrow S$. In our case $S=G$ and I'm looking at a group action defined as the "conjugation". If it depends on the right/left group action isn't this saying a different group action (conjugation) depends on the definition of another group action (the left vs right group action)? – QPhysl Aug 12 '23 at 00:06
  • 4
    The axiom μ(x∗y,s)=μ(x,μ(y,s)) is for a left group action. The difference in the two definitions of conjugation is that one is a left action and the other is a right action. – scleaver Aug 12 '23 at 00:50
  • 3
    For a right action the axiom would be μ(x∗y,s)=μ(y, μ(x, s)). – scleaver Aug 12 '23 at 00:59
  • 2
    The simplest explanation may be that your source has an inconsistency where they defined a group action as a left action but when they defined conjugation, they defined it as a right action. – Ted Aug 12 '23 at 04:03
  • 1
    I agree with Steve Kass, scleaver and Ted who know this at least as well as I do. This is the difference between a left action vs. a right action. About conjugation in particular, many authors use right action simply because they write $x^g$ for the conjugate. With the acting element $g$ on the right, we want the rule to be $(x^g)^h=x^{gh}$, and this, in turn, forces us to define $x^g=g^{-1}xg$ as opposed to $x^g=gxg^{-1}$. FWIW in my lecture notes I do define conjugation as $gxg^{-1}$ because I default to left actions (I'm not a group theorist). Then I cannot use exponential notation. – Jyrki Lahtonen Aug 12 '23 at 06:27
  • 2
    (cont'd) And students still need to learn about both left/right actions. If a group $G$ acts on a set $X$ from the left, and $A$ is any set, then $G$ naturally acts on the set of functions $f:A\to X$ from the left, but the natural action on the set of functions $f:X\to A$ is from the right. Alternatively you can define right action by $g$ to mean the same thing as left action by $g^{-1}$. – Jyrki Lahtonen Aug 12 '23 at 06:35
  • 1
    @JyrkiLahtonen Why not use the notation $^gx$ for $gxg^{-1}$? I usually prefer right actions (mainly because in English and European languages we read from left to right), but when obliged to use left actions I write $^gx$ for conjugation. – Derek Holt Aug 12 '23 at 07:47

0 Answers0