Let g be an element of a group G. If |G| is finite and even, show that g neq 1 in G exists such that g2 = 1

Ayaana Buck 2021-03-02 Answered

Let g be an element of a group G. If \(|G|\) is finite and even, show that \(g \neq 1\) in G exists such that \(g^2 = 1\)

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Expert Answer

Alix Ortiz
Answered 2021-03-03 Author has 9388 answers

Pair up if possible each element of   G with its inverse, and observe that
\(g^{2}\) not equal \(e\Leftrightarrow\) not equal \(g^{−1} \Leftrightarrow\) there exists the pair \((g,g^{−1})\)
Now, there is one element that has no pairing: the identity   e (since indeed   \(e=e−1  \Leftrightarrow  e2=e,e=e −1\) \(\Leftrightarrow e^{2} =e\)), so since the number of elements of   G is even there must be at least one element more, say   e not equal \(a\in G\)  e not equal \(a\in G\), without a pairing, and thus   \(a=a−1  \Leftrightarrow  a2=e.a=a^{−1} \Leftrightarrow a^{2} =e\).

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