Group acts as automorphisms by conjugation

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Let G be a group. For any g \in G, define the map:

c_g:G \to G

given by:

c_g(x) := gxg^{-1}

(this is termed conjugation by g or the inner automorphism induced by g).

Then, the following are true:

Related facts

Analogues in other structures


Proof that every conjugation is an automorphism

Given: A group G, an element g \in G

To prove: The map c_g: x \mapsto gxg^{-1} is an automorphism of G

Proof: c_g is a map from G to G. We need to thus prove three things:

  • c_g(xy) = c_g(x)c_g(y)

The key idea for this is the fact that g^{-1}g = e. Formally:

c_g(x)c_g(y) = (gxg^{-1})(gyg^{-1}) = gx(g^{-1}g)yg^{-1} = gxeyg^{-1} = gxyg^{-1} = c_g(xy)

  • c_g(e) = e

The key idea for this is the fact that gg^{-1} = e. Formally:

c_g(e) = geg^{-1} = gg^{-1} = e

  • c_g(x^{-1}) = c_g(x)^{-1}

Here, we use the fact that the inverse of a product of elements is the product of their inverses, in reverse order (i.e. the inverse map is involutive). Thus, we have:

c_g(x)^{-1} = (gxg^{-1})^{-1} = (g^{-1})^{-1}x^{-1} g^{-1} = gx^{-1}g^{-1} = c_g(x^{-1})

Note that for the proof to go through we need to use the fact that g and g^{-1} are left and right inverses of each other.

NOTE: It actually suffices to prove only the first of these three things, because to test whether a map between groups is a homomorphism of groups, it suffices to check whether it sends products to products. However, when working in somewhat greater generality than groups, it becomes important to check the other conditions, and they're explained here for illustrative purposes.

Thus, every c_g is a homomorphism. It remains to show that this homomorphism is injective and surjective. Injectivity is clear, because c_g(x) = e \implies x = e. For surjectivity, note that given any y \in G, setting:

x := g^{-1}yg

yields c_g(x) = y. (This also becomes clearer from the fact that the map g \mapsto c_g is itself a homomorphism of groups).

Thus, c_g is an automorphism.

Proof that the map is a homomorphism

Given: A group G

To prove: The map g \mapsto c_g, is a homomorphism from G to \operatorname{Aut}(G)

Proof: We'll check three things:

  • c_{gh} = c_g \circ c_h

The proof again uses the reversal law for inverses (inverse map is involutive):

c_{gh}(x) = ghx(gh)^{-1} = ghxh^{-1}g^{-1} = g(c_h(x))g^{-1} = c_g(c_h(x)) = (c_g \circ c_h)(x)

  • c_e is the identity map:

c_e(x) = exe^{-1} = x

  • c_g^{-1} = c_{g^{-1}}

This follows from the first two facts, because c_g \circ c_{g^{-1}} = c_{gg^{-1}} = c_e which is the identity map, and the same holds for composition in the other order.