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

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