Order of Conjugate Element equals Order of Element
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Theorem
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then
- $\forall a, x \in \struct {G, \circ}: \order {x \circ a \circ x^{-1} } = \order a$
where $\order a$ denotes the order of $a$ in $G$.
Corollary
- $\forall a, x \in \struct {G, \circ}: \order {x \circ a} = \order {a \circ x}$
Proof
Let $\order a = k$.
Then $a^k = e$, and:
- $\forall n \in \N_{>0}: n < k \implies a^n \ne e$
by definition of the order of $a$ in $G$
We have:
\(\ds \paren {x \circ a \circ x^{-1} }^k\) | \(=\) | \(\ds x \circ a^k \circ x^{-1}\) | Power of Conjugate equals Conjugate of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ e \circ x^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x \circ x^{-1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e\) |
Thus $\order {x \circ a \circ x^{-1} } \le \order a$.
Now suppose $a^n = y, y \ne e$.
Then:
- $x \circ a^n \circ x^{-1} = x \circ y \circ x^{-1}$
If $x \circ y = e$, then:
- $x \circ a^n \circ x^{-1} = x^{-1}$
If $y \circ x^{-1} = e$, then:
- $x \circ a^n \circ x^{-1} = x$
So:
- $a^n \ne e \implies x \circ a^n \circ x^{-1} = \paren {x \circ a \circ x^{-1} }^n \ne e$
Thus:
- $\order {x \circ a \circ x^{-1} } \ge \order a$
and the result follows.
$\blacksquare$
Sources
- 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 8$: The Order (Period) of an Element: $\text{(iii)}$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 41 \gamma$
- 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 $10.18$