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

Pair up if possible each element of G with its inverse, and observe that g2 not equal e⇔ not equal g−1⇔ there exists the pair (g,g−1) Now, there is one element that has no pairing: the identity e (since indeed e=e−1 ⇔ e2=e,e=e−1 ⇔e2=e), so since the number of elements of G is even there must be at least one element more, say e not equal a∈G e not equal a∈G, without a pairing, and thus a=a−1 ⇔ a2=e.a=a−1⇔a2=e.

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Ayaana Buck

Answered question

2021-03-02

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 $g^{2} = 1$

Answer & Explanation

Alix Ortiz

Skilled2021-03-03Added 109 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 $e = e - 1 \Leftrightarrow e 2 = e, e = e - 1$ $\Leftrightarrow e^{2} = e$ ), 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 $a = a - 1 \Leftrightarrow a 2 = e . a = a^{- 1} \Leftrightarrow a^{2} = e$ .

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