Self-Inverse Elements Commute iff Product is Self-Inverse
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Theorem
Let $\struct {G, \circ}$ be a group.
Let $x, y \in \struct {G, \circ}$, such that $x$ and $y$ are self-inverse.
Then $x$ and $y$ commute if and only if $x \circ y$ is also self-inverse.
Proof
Let the identity element of $\struct {G, \circ}$ be $e_G$.
Necessary Condition
Let $x$ and $y$ commute.
Then:
\(\ds \paren {x \circ y} \circ \paren {x \circ y}\) | \(=\) | \(\ds x \circ \paren {y \circ x} \circ y\) | $\circ$ is associative | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ \paren {x \circ y} \circ y\) | $x$ and $y$ commute | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \circ x} \circ \paren {y \circ y}\) | $\circ$ is associative | |||||||||||
\(\ds \) | \(=\) | \(\ds e_G \circ e_G\) | $x$ and $y$ are self-inverse | |||||||||||
\(\ds \) | \(=\) | \(\ds e_G\) | Definition of Identity Element |
Thus $\paren {x \circ y} \circ \paren {x \circ y} = e_G$, proving that $x \circ y$ is self-inverse.
$\Box$
Sufficient Condition
Now, suppose that $x \circ y$ is self-inverse.
We already have that $x$ and $y$ are self-inverse.
Thus:
- $\paren {x \circ x} \circ \paren {y \circ y} = e_G \circ e_G = e_G$
Because $x \circ y$ is self-inverse, we have:
- $\paren {x \circ y} \circ \paren {x \circ y} = e_G$
Thus:
\(\ds \paren {x \circ y} \circ \paren {x \circ y}\) | \(=\) | \(\ds \paren {x \circ x} \circ \paren {y \circ y} = e_G\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \circ \paren {y \circ x} \circ y\) | \(=\) | \(\ds x \circ \paren {x \circ y} \circ y\) | $\circ$ is associative | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y \circ x\) | \(=\) | \(\ds x \circ y\) | Cancellation Laws |
So $x$ and $y$ commute.
$\blacksquare$
Sources
- 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: Examples: $(4)$
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $4$. Groups: Exercise $14$