Group conjugate

https://arbital.com/p/group_conjugate

by Patrick Stevens Jun 16 2016 updated Jun 20 2016

Conjugation lets us perform permutations "from the point of view of" another permutation.


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$~$ %%note:Which 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.

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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.

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 [group_centraliser 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. %%%