 # 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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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   $$a=a−1 \Leftrightarrow a2=e.a=a^{−1} \Leftrightarrow a^{2} =e$$.