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

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 $σ = (a_{11} a_{12} \dots a_{1 n_{1}}) (a_{21} \dots a_{2 n_{2}}) \dots (a_{k 1} a_{k 2} \dots a_{k n_{k}})$ and $τ \in S_{n}$ , then $τ σ τ^{- 1} = (τ (a_{11}) τ (a_{12}) \dots τ (a_{1 n_{1}})) (τ (a_{21}) \dots τ (a_{2 n_{2}})) \dots (τ (a_{k 1}) τ (a_{k 2}) \dots τ (a_{k n_{k}}))$

That is, conjugating by $τ$ has "caused us to view $σ$ from the point of view of $τ$ ".

Similarly, in the dihedral group $D_{2 n}$ 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 $X$ ^[1] then conjugation of $g \in G$ by $h \in G$ produces a symmetry $h g h^{- 1}$ 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.

Conjugation action

Conjugation forms a action. Formally, let $G$ act on itself: $ρ : G \times G \to G$ , with $ρ (g, k) = g k g^{- 1}$ . It is an exercise to show that this is indeed an action.

Show solution

We need to show that the identity acts trivially, and that products may be broken up to act individually.

$ρ (g h, k) = (g h) k (g h)^{- 1} = g h k h^{- 1} g^{- 1} = g ρ (h, k) g^{- 1} = ρ (g, ρ (h, k))$ ;

$ρ (e, k) = e k e^{- 1} = k$ .

The stabiliser of this action, ${S t a b}_{G} (g)$ for some fixed $g \in G$ , is the set of all elements such that $k g k^{- 1} = g$ : that is, such that $k g = g k$ . 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, ${O r b}_{G} (g)$ 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.

^{^︎}
Which we can always view as being the case.