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Set of all elements of a given order is not necessarily a subgroup

Let $n \geq 3$. Show that $\{ x \in D_{2n} \ |\ x^2 = 1\}$ is not a subgroup of $D_{2n}$.

Solution: We know that every element of $D_{2n}$ which is not a power of $r$ has order 2; in particular, $sr$ and $sr^2$ have order 2. But $$srsr^2 = ssr^{-1}r^2 = r$$ has order $n \geq 3$. Thus this set is not closed under multiplication.


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