Group conjugate

Two el­e­ments \(x, y\) of a group \(G\) are con­ju­gate if there is some \(h \in G\) such that \(hxh^{-1} = y\).

Con­ju­gacy as “chang­ing the wor­ld­view”

Con­ju­gat­ing by \(h\) is equiv­a­lent to “view­ing the world through \(h\)’s eyes”. This is most eas­ily demon­strated in the sym­met­ric group, where it is a fact that if \($\sigma = (a_{11} a_{12} \dots a_{1 n_1})(a_{21} \dots a_{2 n_2}) \dots (a_{k 1} a_{k 2} \dots a_{k n_k})\)$ and \(\tau \in S_n\), then \($\tau \sigma \tau^{-1} = (\tau(a_{11}) \tau(a_{12}) \dots \tau(a_{1 n_1}))(\tau(a_{21}) \dots \tau(a_{2 n_2})) \dots (\tau(a_{k 1}) \tau(a_{k 2}) \dots \tau(a_{k n_k}))\)$

That is, con­ju­gat­ing by \(\tau\) has “caused us to view \(\sigma\) from the point of view of \(\tau\)”.

Similarly, in the dihe­dral group \(D_{2n}\) on \(n\) ver­tices, con­ju­ga­tion of the ro­ta­tion by a re­flec­tion yields the in­verse of the ro­ta­tion: it is “the ro­ta­tion, but viewed as act­ing on the re­flected poly­gon”. Equiv­a­lently, if the poly­gon is sit­ting on a glass table, con­ju­gat­ing the ro­ta­tion by a re­flec­tion makes the ro­ta­tion act “as if we had moved our head un­der the table to look up­wards first”.

In gen­eral, if \(G\) is a group which acts as (some of) the sym­me­tries of a cer­tain ob­ject \(X\) noteWhich we can always view as be­ing the case. then con­ju­ga­tion of \(g \in G\) by \(h \in G\) pro­duces a sym­me­try \(hgh^{-1}\) which acts in the same way as \(g\) does, but on a copy of \(X\) which has already been per­muted by \(h\).

Clo­sure un­der conjugation

If a sub­group \(H\) of \(G\) is closed un­der con­ju­ga­tion by el­e­ments of \(G\), then \(H\) is a nor­mal sub­group. The con­cept of a nor­mal sub­group is ex­tremely im­por­tant in group the­ory.

knows-req­ui­site(Group ac­tion):

Con­ju­ga­tion action

Con­ju­ga­tion forms a ac­tion. For­mally, let \(G\) act on it­self: \(\rho: G \times G \to G\), with \(\rho(g, k) = g k g^{-1}\). It is an ex­er­cise to show that this is in­deed an ac­tion. %%hid­den(Show solu­tion): We need to show that the iden­tity acts triv­ially, and that prod­ucts may be bro­ken up to act in­di­vi­d­u­ally.

  • \(\rho(gh, k) = (gh)k(gh)^{-1} = ghkh^{-1}g^{-1} = g \rho(h, k) g^{-1} = \rho(g, \rho(h, k))\);

  • \(\rho(e, k) = eke^{-1} = k\). %%

The sta­bil­iser of this ac­tion, \(\mathrm{Stab}_G(g)\) for some fixed \(g \in G\), is the set of all el­e­ments such that \(kgk^{-1} = g\): that is, such that \(kg = gk\). Equiv­a­lently, it is the cen­tral­iser of \(g\) in \(G\): it is the sub­group of all el­e­ments which com­mute with \(G\).

The or­bit of the ac­tion, \(\mathrm{Orb}_G(g)\) for some fixed \(g \in G\), is the con­ju­gacy class of \(g\) in \(G\). By the or­bit-sta­bil­iser the­o­rem, this im­me­di­ately gives that the size of a con­ju­gacy class di­vides the or­der of the par­ent group. <div>


  • Group

    The alge­braic struc­ture that cap­tures sym­me­try, re­la­tion­ships be­tween trans­for­ma­tions, and part of what mul­ti­pli­ca­tion and ad­di­tion have in com­mon.