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I am trying to solve the following exercise about the dihedral group and its center:

If $g\in Z(D_{2n})\Leftrightarrow ga=ag, bg=gb$, where $a,b$ are generators of $D_{2n}$.

We have defined the dihedral group of order $2n$ as $D_{2n}=\left \{ a^{i}b^{j} \mid 0\leq i \leq n, j=0,1 \right \}$, the group of all finite products of the generators $a$ and $b$, where $a= \bigl(\begin{smallmatrix} 1 & 2 & 3 & \cdots & n-1 & n \\ \end{smallmatrix}\bigr)$ and $ b = \bigl(\begin{smallmatrix} 1 & 2 & 3 & \cdots & n-1 & n \\ 1 & n & n-1 & \cdots & 3 & 2 \end{smallmatrix}\bigr)$

For the first part ( $\Rightarrow $) i have considered an element $g$ of the center $Z(D_{2n})$, $g$ is also in $D_{2n}$, so it can be written as a finite product of powers of $a$ and $b$. Let's say $g=a^{k}b^{l}$ for $0 \leq k \leq n$ and $j=0$ or $j=1$. Then i can write $ga = a^{k}b^{l}a=a^{k+1}b^{l}$ and $ag=aa^{k}b^{l}=a^{k+1}b^{l}$. So $ga=ag$. The same for $gb=bg$.

Is this correct? How to start with the other part of the proof? I would appreciate any help and hints. Thank you in advance.

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I don't think I understood what you wrote. Look, if $g\in Z(D_{2n})$, then $gh=hg$ for every $h\in D_{2n}$. In particular it is true for $h=a$ and $h=b$. This proves the half you stated above.

Conversely, suppose $ga=ag$ and $gb=bg$. First note that $ga^i=a^ig$ for all $1\leq i<n$ (you can prove this by induction if you like: $ga^{i}=aga^{i-1}$). Now, any $h\in D_{2n}$ has the form $h=a^ib^j$, $0\leq i<n$, $0\leq j\leq 1$. So, $$gh=ga^ib^j=a^igb^j=a^ib^jg=hg.$$ This completes the proof.

This gives a fast way to prove that $Z(D_{2n})$ is nontrivial only if $n=2k$ is even. In this case, either $k=1$ and $D_4=\mathbb{Z}_2\times\mathbb{Z}_2$ is abelian, or $Z(D_{2n})=\langle a^k\rangle$. To see this, suppose $g\in Z(D_{2n})\backslash\{1\}$. We consider two cases:

Case 1: $g=a^k$. Then $g$ commutes with $a$. Next we compute $gb=a^kb$, and $bg=ba^k=a^{-k}b$. Since $bg=gb$, we must have $a^{k}b=a^{-k}b$. This forces $k\equiv -k$ (mod $n$), so $n=2k$ is even.

Case 2: $g=a^kb$. Then $ag=a^{i+1}b$ and $ga=a^iba=a^{i-1}b$, so $i+1\equiv i-1$ (mod $n$). This forces $n=2$, and $D_{2n}=D_4=\mathbb{Z}_2\times\mathbb{Z}_2$.

Putting these cases together, $Z(D_{2n})=1$ if $n$ is odd; if $n=2$, $D_4=Z(D_4)$ is abelian; if $n=2k>2$, then $Z(D_{2n})=\langle a^k\rangle$.

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  • $\begingroup$ Thank you very much, this is great! $\endgroup$ – Lullaby Oct 7 '15 at 20:35
  • $\begingroup$ @DavidHill does this answer only pertain to even dihedral groups, because of your subscript $2n$? I'm confused... $\endgroup$ – ALannister Jan 1 '17 at 22:49
  • $\begingroup$ @JessyCat $D_{2n}$ is the symmetry group of a regular $n$-gon. It has order $2n$. $\endgroup$ – David Hill Jan 2 '17 at 4:29

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