Translation and Conjugation

Translation and conjugation

Definition (Translation)

Let \((G, \ast)\) be a group and let \(a \in G\). The left (right) translation by \(a\) is \(T_{a}:G\to G\) such that \(x \mapsto a \ast x\) (\(x \mapsto x \ast a\)).

Note (Translation is bijective)

1. \(T_{a}\) is injective, since \(T_{a}(x)=T_{a}(y) \iff a \ast x = a \ast y \iff x = y\).
2. \(T_{a}\) is surjective, since if \(y \in G\), \(T_{a}(a^{-1} \ast y) = y\).

Hence \(T_{a}\) is bijective. Conclude that \(T_{a} \in S_{G} = \{ \sigma:G\to G \text{ bijective}\}\).

Theorem (Translations of \(G\) are a monomorphism)

Let \(f:G\to S_{G}\) be such that \(f(a) = T_{a}\). Then \(f\) is a monomorphism.

Proof: We first show that \(f\) is a homomorphism, i.e., that \(f(a \ast b) = T_{a \ast b} = T_{a} \ast T_{b}\). Let \(x \in G\). We have that

\[\begin{aligned} (T_{a} \circ T_{b})(x) = T_{a}(T_{b}(x)) &= T_{a} (b \ast x)\\ &= a \ast (b \ast x) \\ &=(a \ast b) \ast x \\ &= T_{a \ast b} (x) \end{aligned},\]

from which we have that \(f\) is a homomorphism. We now show that \(f\) is injective. Let \(a \in \operatorname{Ker}(f)\). Let \(a \in G\) be such that \(f(a) = T_{a} = \mathrm{id}_{G}\). Hence,

\[\forall x \in G \quad(T_{a}(x)= x) \implies \forall x \in G \quad(a \ast x = x) \implies a = 1_{G}.\]

Thus \(\operatorname{Ker}(f) = \{ \mathrm{id_{G}} \}\). We conclude that \(f\) is a monomorphism.

Definition (Conjugation)

Let \(G\) be a group and \(a \in G\). Define conjugation by \(a\) \(C_{a}: G\to G\) such that \(x \mapsto a \ast x \ast a^{-1}\).

Theorem (Conjugation in the automorphism)

Let \(G\) be a group and \(a \in G\). Then \(C_{a} \in \mathrm{Aut}(G)\).

Proof: Note that \(C_{a}(x \ast y) = C_{a}(x) \ast C_{a}(y)\) (exercise), so \(C_{a}\) is a homomorphism. To show that it is injective, take \(x \in \operatorname{Ker}(C_{a})\). Hence

\[C_{a}(x) = 1_{G} \implies a \ast x \ast a ^{-1} = 1_{G} \implies x = 1_{G}\]

To show that it is surjective, let \(x \in G\). Note that \(C_{a}(a^{-1} \ast x \ast a) = x\), from which we conclude surjectivity.

Theorem (Conjugations of \(G\) are a homomorphism)

Let \(f:G\to \mathrm{Aut}(G) \leq S_{G}\) be such that \(f(a) = C_{a}\). Then \(f\) is a homomorphism.

Proof: Let \(x \in G\). Then

\[\begin{aligned} (C_{a} \circ C_{b})(x) &= C_{a} (C_{b}(x))\\ &=C_{a} ( b \ast x \ast b^{-1})\\ &=a \ast ( b \ast x \ast b^{-1}) \ast a ^{-1} \\ &=(a \ast b) \ast x \ast (a \ast b) ^{-1} \\ &=C_{a \ast b}(x). \end{aligned}\]