# 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.

• $\rho(gh, k) = (gh)k(gh)^{-1} = ghkh^{-1}g^{-1} = g \rho(h, k) g^{-1} = \rho(g, \rho(h, k))$;
• $\rho(e, k) = eke^{-1} = k$. %%

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 [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. %%%