Power of Conjugate equals Conjugate of Power
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
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 6.6$. Normal subgroups: Example $124 \ \text{(ii)}$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: An Introduction to Groups: Exercise $12$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Exercise $7$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.18$