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

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

• just look that $s_1 \times s_2 = r_1$ if $s_1 \neq s_2$ – Just do it Apr 13 at 6:15
• What is $D_n$ for you? Many sources (the ones I trust in particular) define $D_n$ to be the group of symmetries of a regular $n$-gon. Some other sources, OTOH, call that group $D_{2n}$. I guess their motivation is that the subscript would be the order of the group. I'm not buying that, but to each book author their own, I suppose :-) – Jyrki Lahtonen Apr 13 at 7:00

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 $$s=s^{-1}$$

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

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

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 $$S$$.

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.