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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?

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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
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1 Answer

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.

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