0

Let $n \geq 3.$ Show that $Z(D_n) \cong \Bbb Z_2$ if $n$ is even, and $Z(D_n)=\{e\}$ when $n$ is odd.

First, let $n$ be odd. Assume for the contrary that there exists $\sigma \in D_n$ such that $e \neq \sigma$ and $\sigma \tau=\tau \sigma$ for all $\tau \in D_n$, where $e$ denotes the identity in $D_n.$ Since $\sigma \neq e$, there exists $i,j$ positive distinct integers such that $\sigma(i)=j.$ Let $k \neq i,j$ and let $\tau=(i k)$\ (such a $k$ exists since $n \geq 3$). By our assumption, we should have $\sigma \tau=\tau \sigma$, so $$\sigma(\tau(i))=\sigma(k)\neq j=\tau(j)=\tau(\sigma (i))$$ a contradiction, so $\sigma$ must equals $e$. Hence, $Z(D_n)=\{e\}$ i.e the $Z(D_n)$ is trivial, for $n$ odd. I feel i am ok with the part when $n$ is odd, but i am getting stuck when the case $n$ is even. So I would really appreciate any help or thoughts about that.

An alternative proof:

assume that $e \neq g \in Z(D_n)$. Then, $g=a^ib^j$ where $1 \leq i \leq n,\ \ j \in \{0,1\}$, so we have two cases:\ case (i) $g=a^i$. By the assumption that $ g \in Z(D_n)$ and the relation between $a,b$, we have $a^ib=ba^i=a^{-i}b$, therefore $a^i=a^{-i}$ which implies that $e^{2i}=1$. So $e^{2i}=1=e^n$ which means $i=\frac{n}{2}.$\ case(ii) $g=a^ib$. By assumption and the relation between $a,b$,$$a^i=(a^ib)b=b(a^ib)=(ba^i)b=a^{-i}bb=a^{-i}$$ so $a^{i}=a^{-i}$, which implies $i=\frac{n}{2}$. Also, for $a$ we have $$(a^ib)a=a^i(ba)=a^{i-1}b=a(a^ib)=a^{i+1}b$$ i.e $a^{i-1}=a^{i+1}$ here we have $a^2=1$, so what does that mean? and how to show $D_n \cong Z_2$ when $n=2k.$

Thank you.

Ahmed
  • 1,486
  • 1
    I assume that the element in the center of $D_n$ is $\sigma^{\frac n2} \tau$. Does that help? – Robert Shore Sep 25 '19 at 01:46
  • 1
    Hint: Can you find out the elements of the center? Take a guess. Show that they are in center and others are not (using generator-relation). – nobody Sep 25 '19 at 01:46
  • 2
    Alternatively, you can just write out the equations systematically since you know all the elements in terms of the two generators. – nobody Sep 25 '19 at 01:48
  • 1
    This has been asked here before, I just don't have my answers organized well enough to find it easily. –  Sep 25 '19 at 02:01
  • 1
    Found it: https://math.stackexchange.com/questions/1469131/center-of-dihedral-group – Travis Willse Sep 25 '19 at 02:43

1 Answers1

2

Hint As usual, let $r$ denote a generator of the subgroup of rotations and $s$ any reflection. Then, we can write any element of $D_n$ as $r^k s^l$ for $0 \leq k < n$ and $l \in \{0, 1\}$. The facts $r^n = s^2 = e$ and $sr = r^{-1} s$ are enough to rewrite the product $(r^k s^l) (r^{k'} s^{l'})$ in the form $r^p s^q$.

Travis Willse
  • 99,363
  • Please take a look, I change the entire proof – Ahmed Sep 25 '19 at 14:37
  • You're essentially done, you just have to interpret your computations: You've shown that for an element of the form $r^k$ to be in the center, we must have $r^{2 k} = 1$. So, if $n$ is even, the only such $k$ are (modulo $n$) $0, \frac{n}{2}$. You've also shown that for an element of the form $r^k s$ to be in the center it is necessary and sufficient for $a^2 = 1$. But that implies that $n | 2$, so the condition $n \geq 3$ specified in the problem is not satisfied. Thus, $Z(D_n)$, $n \geq 3$ and $n$ even, has two elements and hence $Z(D_n) \cong \Bbb Z_2$. – Travis Willse Sep 25 '19 at 15:02
  • Of course, $D_1$ and $D_2$ are abelian (they have fewer than $6$ elements), so $Z(D_1) = D_1 \cong \Bbb Z_2$ and $Z(D_2) = D_2 \cong \Bbb Z_2 \times \Bbb Z_2$, which do not fit the identified pattern for $n \geq 3$. – Travis Willse Sep 25 '19 at 15:04
  • I think i must include the case when $g=b?$ dont you think so? – Ahmed Sep 25 '19 at 16:48
  • That's included in your case II, where you consider elements of the form $a^i b$. If $i = 0$ then $a^i b = b$. – Travis Willse Sep 25 '19 at 16:55
  • Ah, I think i made a mistake when saying $1 \leq i \leq n$, it should be $0 \leq i \leq n$ – Ahmed Sep 25 '19 at 17:15