Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $C_G(x)$ be the centralizer of $x$, $G=D_{2p}$ with $2p$ elements the dihedral group and $C_p$ the cyclic group. I'm looking for an element $x\in G$, that does NOT hold the equation $$C_G(x \bmod C_p ) = C_G(x) C_p $$ Can anyone help me please?

share|improve this question
You cannot compute a centralizer in $G$ of an element of $G/C_p$, so $C_G(x\bmod C_p)$ does not really make sense. Do you mean, $C_{G/C_p}(x\bmod C_p)$? And is $p$ necessarily a prime? An odd prime? –  Arturo Magidin Apr 19 '12 at 19:31
I'm using the text ams.org/mathscinet-getitem?mr=276318 and try to get, that the inequation $k(G)<k(G/N)k(N)$. For this I need that $C(σ modN)=C(σ)N $ is wrong for one element of $G$. The notations are the same as in the article. And yes, $p$ is an odd prime. –  amy Apr 19 '12 at 19:39
Doesn't really matter what text you are using, what you wrote makes no sense for the reason I explained: you cannot compute the centralizer in $G$ of an element in $G/C_p$, which is neither a subgroup nor an overgroup of $G$. The mathreview in question is abusing notation, probably to mean something along the lines of what I wrote. And what about $p$? –  Arturo Magidin Apr 19 '12 at 19:41
Sorry, I wasn't ready with my comment –  amy Apr 19 '12 at 19:44
Ok, would it make sense, if the first centralizer would be $C_{G/C_p}$? –  amy Apr 19 '12 at 19:47

1 Answer 1

up vote 1 down vote accepted

Let $D_{2n} = \langle r,s\mid r^n = s^2 = 1,\ sr=r^{-1}s\rangle.$

We have, with $0\leq i,t\lt n$, $0\leq j,k \lt 2$, $$(r^is^j)(r^ts^k) = \left\{\begin{array}{cc}r^{(i-t)\bmod n}s^{(j+k)\bmod 2}&\text{if }j=1\\ r^{(i+t)\bmod n}s^{(j+k)\bmod 2}&\text{if }j=0.\end{array}\right.$$ So:

  1. If $j=k=0$, then $r^ir^t = r^tr^i$.
  2. If $j=0$, $k=1$, then $r^i(r^ts) = (r^ts)r^i$ if and only if $t-i\equiv t+i\pmod{n}$, if and only if $2i\equiv 0\pmod{n}$, if and only if $i=0$ or $n$ is even and $i=n/2$.
  3. If $k=1$, $j=0$, then $(r^is) r^t = r^t(r^is)$ if and only if $i-t\equiv i+t\pmod{n}$, if and only if $2t\equiv 0\pmod{n}$, if and only if $t=0$, or $n$ is even and $t=n/2$.
  4. If $j=k=1$, then $(r^is)(r^ts) = (r^ts)(r^is)$ if and only if $i-t\equiv t-i\pmod{n}$, if and only if $2t\equiv 2i\pmod{n}$, if and only if $i\equiv t\pmod{n/\gcd(n,2)}$.

So: since you are working with $n=p$ an odd prime, we have:

  1. If $i\neq 0$, $C_G(r^i) = \langle r\rangle$. All these elements work: $C_G(r^i) = C_p$, $r^i\bmod C_p = C_p$, $C_{G/C_p}(r^i\bmod C_p) = C_{G/C_p}(C_p) = G/C_p$, but $C_G(r^i)C_p$ is the trivial element of $G/C_p$.

  2. $C_G(s) = \langle s\rangle$. Here, $C_{G/C_p}(s\bmod C_p) = G/C_p$, but $C_G(s)C_p = \langle s\rangle C_p = G/C_p$ as well.

  3. If $i\neq 0$, $C_g(r^is) = \langle r^is\rangle$. Here $C_{G/C_p}(r^is\bmod C_p)=G/C_p$, and $C_{G}(s)C_p = \langle r^is\rangle C_p = G/C_p$ as well.

So any nontrivial power of $r$ will work.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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