3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ just look that $s_1 \times s_2 = r_1$ if $s_1 \neq s_2$ $\endgroup$ – Just do it Apr 13 at 6:15
  • 2
    $\begingroup$ 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 :-) $\endgroup$ – Jyrki Lahtonen Apr 13 at 7:00
2
$\begingroup$

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 $s=s^{-1}$

$\endgroup$
3
$\begingroup$

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 $2$

$\endgroup$
1
$\begingroup$

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 $S$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.