Power of Conjugate equals Conjugate of Power

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Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $x, y \in G$ such that $\exists a \in G: x \circ a = a \circ y$.

That is, let $x$ and $y$ be conjugate.


Then:

$\forall n \in \Z: y^n = \paren {a^{-1} \circ x \circ a}^n = a^{-1} \circ x^n \circ a$

It follows directly that:

$\exists b \in G: \forall n \in \Z: y^n = b \circ x^n \circ b^{-1}$


In particular:

$y^{-1} = \paren {a^{-1} \circ x \circ a}^{-1} = a^{-1} \circ x^{-1} \circ a$


Proof

Proof by induction:

For all $n \in \N$, let $\map P n$ be the proposition $y^n = a^{-1} \circ x^n \circ a$.


$\map P 0$ is true, as this just says $e = a^{-1} \circ e \circ a$.


Basis for the Induction

$\map P 1$ is the case $y = a^{-1} \circ x \circ a$, which is how conjugacy is defined for $x$ and $y$.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$y^k = a^{-1} \circ x^k \circ a$


Then we need to show:

$y^{k + 1} = a^{-1} \circ x^{k + 1} \circ a$


Induction Step

This is our induction step:

\(\ds y^{k + 1}\) \(=\) \(\ds \paren {a^{-1} \circ x \circ a}^{k + 1}\)
\(\ds \) \(=\) \(\ds \paren {a^{-1} \circ x \circ a}^k \circ \paren {a^{-1} \circ x \circ a}\)
\(\ds \) \(=\) \(\ds \paren {a^{-1} \circ x^k \circ a} \circ \paren {a^{-1} \circ x \circ a}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds a^{-1} \circ x^k \circ x \circ a\)
\(\ds \) \(=\) \(\ds a^{-1} \circ x^{k + 1} \circ a\)


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \N: y^n = a^{-1} \circ x^n \circ a$


Now we need to show that if $\map P n$ holds, then $\map P {-n}$ holds.

That is:

$y^{-n} = a^{-1} \circ x^{-n} \circ a$


Let $n \in \N$.

Then:

\(\ds y^{-n}\) \(=\) \(\ds \paren {a^{-1} \circ x \circ a}^{-n}\)
\(\ds \) \(=\) \(\ds \paren {\paren {a^{-1} \circ x \circ a}^n}^{-1}\) Index Laws for Monoids: Negative Index
\(\ds \) \(=\) \(\ds \paren {a^{-1} \circ x^n \circ a}^{-1}\)
\(\ds \) \(=\) \(\ds a^{-1} \circ \paren {x^n}^{-1} \circ \paren {a^{-1} }^{-1}\) Inverse of Group Product
\(\ds \) \(=\) \(\ds a^{-1} \circ x^{-n} \circ a\)


Thus $\map P n$ has been shown to hold for all $n \in \Z$.

Hence the result.

$\blacksquare$


Sources

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