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This is a homework problem I got. My approach is: if $n$ is odd, then $D_n=\{1,r,r^2,...,r^{n-1},s,rs,...,r^{n-1}s\}$, since $sr=r^{n-1}s$ (not equal to $rs$), and this generalizes to all compositions of $s$ and $r$ (because $n$ is odd), the only element in $Z(D_n)$ is $1$.

But I have trouble with the second part of the proof. $\mathbb Z_2=\{0,1\}$, it's not even a subset of $D_n$, so I don't understand what they are asking here.

Shaun
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Scharfschütze
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    With regard to $\mathbb{Z}_2$ not being a subset of $D_n$, all that is meant is that when $n$ is even the center of $D_n$ has just two elements, and it is easy to see that any two element group must be cyclic and hence commutative, and so is isomorphic to $\mathbb{Z}_2$ in a natural way. – Rankeya Nov 03 '11 at 02:59
  • hint: one way is to write out the conjugacy classes of Dn for the even and odd cases. What do you know about the relation between the size of a conjugacy class and the centre? – Ralth Nov 03 '11 at 03:12

1 Answers1

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First of all, what they're asking you to show is not that the center is $\mathbb Z_2$, but rather that it is isomorphic to it, which basically means in the language of group theory that they are the "same" in a well-defined sense. If you don't know what isomorphic means I believe your notes probably will explain it with a few pages of details so I'll skip that part.

Now since you know $sr = r^{n-1}s$ in $D_n$, say for instance a $k^{th}$ power of $r$ would be in the center, say $r^k$. Clearly $r^k$ commutes with every power of $k$, but it commutes with $s$ if and only if $sr^k = r^ks$, but $sr^k = r^{n-k}s$ (you can easily show this equality by induction on $k$), so that $r^k$ commutes with $s$ if and only if $2k = n$, which means if and only if $n$ is even and $k = n/2$. Now $s$ cannot be in the center since $sr \neq rs$, unless $n=2$, but I guess you include that case only if you're bored, because usually $D_n$ also refers to symmetries of a polygon, and no polygon has $2$ vertices (besides the line segment.. bleh.) Anyway. My point is, $r^{n/2}$ is in the center, and $s$ is not. Note that similar arguments show that $r^{\ell}s$ is not in the center for any integer $\ell$.

This means that $$ Z(D_n) = \begin{cases} \{1\} & \text{ if } n \equiv 1 (\mathrm{mod} 2) \\ \{1, r^{n/2}\} & \text{ if } n \equiv 0 (\mathrm{mod} 2) \end{cases} $$ but since $\{1, r^{n/2} \}$ is isomorphic to $\mathbb Z_2$, we say that the center "is" $\mathbb Z_2$ in the sense of isomorphisms.

Hope that helps,

Patrick Da Silva
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  • The things I put in capital letters I wanted them in italic. I tried using the \textit command but it didn't work... can anyone tell me what's wrong with that? Thanks in advance. – Patrick Da Silva Nov 03 '11 at 03:33
  • @JackSchmidt : Thanks a lot man – Patrick Da Silva Nov 03 '11 at 03:37
  • I think you combined two statements. "if and only if $sr^k=r^{n−k}s$" should probably be "if and only if $r^k = r^{n-k}$, since $sr^k=r^{n−k}s$". – Jack Schmidt Nov 03 '11 at 03:54
  • I didn't combine two statements, it was just a typo. Thanks for noticing =) – Patrick Da Silva Nov 03 '11 at 03:56
  • All looks good. D2 = 2×2 is a counterexample to the original question, but it is a bit degenerate anyways. I guess D1 = 2 is too, but that is really degenerate. We talk about them as symmetry groups though: D1 is a single reflection, and D2 is generated by two perpendicular reflections. – Jack Schmidt Nov 03 '11 at 03:59