Power of Product of Commutative Elements in Group
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Theorem
Let $\struct {G, \circ}$ be a group.
Let $a, b \in G$.
Then:
- $a \circ b = b \circ a \iff \forall n \in \Z: \paren {a \circ b}^n = a^n \circ b^n$
That is:
- $a$ and $b$ commute
This can be expressed in additive notation in the group $\struct {G, +}$ as:
- $a + b = b + a \iff \forall n \in \Z: n \cdot \paren {a + b} = \paren {n \cdot a} + \paren {n \cdot b}$
Proof
Necessary Condition
Let $a \circ b = b \circ a$.
By definition, all elements of a group are invertible.
Therefore the results in Power of Product of Commutative Elements in Monoid can be applied directly.
$\Box$
Sufficient Condition
If $\paren {a \circ b}^n = a^n \circ b^n$ for all $n$, then it certainly holds for $n = 2$:
\(\ds \paren {a \circ b}^2\) | \(=\) | \(\ds a^2 \circ b^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a \circ b} \circ \paren {a \circ b}\) | \(=\) | \(\ds \paren {a \circ a} \circ \paren {b \circ b}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a^{-1} \circ a} \circ b \circ a \circ \paren {b \circ b^{-1} }\) | \(=\) | \(\ds \paren {a^{-1} \circ a} \circ a \circ b \circ \paren {b \circ b^{-1} }\) | Group Axiom $\text G 1$: Associativity | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds e \circ b \circ a \circ e\) | \(=\) | \(\ds e \circ a \circ b \circ e\) | Group Axiom $\text G 3$: Existence of Inverse Element | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds b \circ a\) | \(=\) | \(\ds a \circ b\) | Group Axiom $\text G 2$: Existence of Identity Element |
$\blacksquare$
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $12$: Applications of Sylow Theory: $(2)$ Groups of order $21$: Proposition $12.2$