Elements of order 2 in $D_{2n}$

Question

Im new at this abstract algebra stuff and im not comfortable with the proofs techniques yet, so I have a question related to the elements of order $2$ in $D_{2n}$.

Problem:

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

By now, I know that all the elements of the form $sr^{k}$ has order $2$ because I tested by brute force haha, but my real question is how do I prove prove that. How do I have to proceed to prove that $sr^k$ has order 2.

Any help will be really appreciated, thanks so much <3

Answer 1

If n=2 the statement is false, as $D_4 \simeq \mathbb{Z}/ 2\mathbb{Z} \times \mathbb{Z}/ 2\mathbb{Z}$ comprises 3 elements of order 2 and the identity. I claim the statement is true for $n\geq 3$. We wish to find $x,y \in D_{2n}$ so that $x^2=y^2=1$ but $(xy)^2\neq 1$. Then the given set is not multiplicatively closed hence not a group. let $r$ and $s$ be elements of $D_{2n}$ satisfying $s^2=r^n=1 , srs=r^{-1}$. Then $s^2=(sr)^2=1$, but if $n\geq 3$, $(s\cdot sr)^2 = r^2 \neq 1$

That $(sr^k)^2=1$ is easy: Observe that $(sr^k)^2 = (sr^k s)r^k = (srs)^k r^k = r^{-k}r^k = 1$, where the second equality follows from the fact the $s=s^{-1}$

Answer 2

Hint as well as Exercise:

If $H$ is a subgroup of $D_{2n}$, then every element of $H$ is a rotation or exactly half of the members of $H$ are rotations

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Note that every reflection has order $2$ and, in addition,if $n$ is even, then there is exactly one rotation has order $2$

Answer 3

Note that the identity $1$ satisfies $1^2 = 1$ and, if $n$ is even, $r^{n / 2}$ satisfies $(r^{n / 2})^2 = r^n = 1$. You should be able to convince yourself that these elements, together with the reflections $s r^k$ you mentioned in the question statement, exhaust all of the elements $x \in D_{2 n}$ such that $x^2 = 1$.

Hint To prove that $S := \{x \in D_{2 n} : x^2 = 1\}$ is not a subgroup, it's enough to find two elements $x, y \in S$ such that $x y$ does not satisfy $(x y)^2 = 1$---as then $S$ would not be closed under the group operation.

NB if $x \in S$ is in the center of $D_{2n}$ then $(xy)^2 = (y x)^2 = x^2 y^2 = 1$, so for any counterexample, $x, y$ cannot be contained in the center of $S$. Moreover, since $D_2, D_4$ are abelian, this observation shows that the claim that $S$ is not a subgroup is false for $n = 1, 2$.

Additional hint Since the center of $D_{2n}$ (for $n > 2$) consists exactly of $1$ and (if $n$ even) $r^{n / 2}$, this leaves only the possibilities $x = s r^k$, $y = s r^\ell$ for some $k, \ell$. Of course, any counterexample must have $k \neq \ell$. So, compute $(s r^k)(s r^\ell)$, and see if you can choose $k, \ell$ for which the product is not in $S$.

Answer 4

It is not a subgroup for $n\gt2$.

Note that $D_{2n}$ has presentation $\langle a,b\mid a^n, b^2, (ba)^2\rangle $.

But, $b(ba)=a$ and $a^2\neq1$. Thus we don't have closure under multiplication.