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I am trying to solve the following problem:

(a) State a formula relating orders of centralizers and cardinalities of conjugacy classes in a finite group $G$.

(b) Let $G$ be a finite group with a proper normal subgroup $N$ that is not contained in the center of $G$. Prove that $G$ has a proper subgroup $H$ with $|H|>|G|^{\frac{1}{2}}$.

For part (a) I have:

Let $x \in G$. Define $f: G/C_G(x) \to C_x$ by $f(gC_G(x))=gxg^{-1}$. Note that $f$ is well defined since if $gC_G(x)=hC_G(x)$ then $h^{-1}g C_G(x)=C_G(x)$ which means that $h^{-1}g \in C_G(x)$, so $h^{-1}gx=xh^{-1}g$ that is $gxg^{-1}=hxh^{-1}$, thus $f(g)=f(h).$ The same reasoning in opposite way shows that $f$ is injective. Moreover $f$ is trivially onto, then $f$ is a bijection.

Therefore $|G/C_G(x)|=[G:C_G(x)]=\frac{|G|}{|C_G(x)|}=|C_x|.$

For part (b) I have.

Since $N$ is proper and it is not contained in $Z(G)$, then let $x \in N-Z(G)$, then by part (a) and by taking $H=C_G(x)$ we have $H$ is a proper subgroup of $G$ as if $H=G$ we have $x \in Z(G).$ Moreover, by our formula in part (a) we have $|G|=|H||C_x|.$

Now I am trying to show that $|C_x|$ is strictly less than $|H|$, then we are done. I know that $|C_x|$is less then $|N|$ because since $N$ is normal to $G$ and $x \in N$ we have $gxg^{-1} \in N$ for all $g \in G$. Then I have $|C_x|<|N|$.

Can anyone give me some hint to conclude that $|C_x|<|H|$? I would like to think a little bit more about it, so please dont give me the direct solution, just some hint for me give the next step and conclude the thing by myself.

Let me know, also, if there is some mistake in my solution for part (a) and/or the solution for part (b) so far.

Thank you so so much for your help!

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As you noticed $|H||C_x|=|G|$ where $H=C_G(x)$. If $|H|>|G|^{1\over2}$, noting to prove. Otherwise, $|N|>|G|^{1\over2}$. In both case, $G$ has a desired proper subgroup.

Note that in $a)$, you should write $f(gC_G(x))=gxg^{-1}$. Rest seems to be true in a).

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  • $\begingroup$ Thank you @mesel... I just did not understand why if $|N|>|G|^{1/2}$ we are done. :( $\endgroup$
    – bttmbrcelo
    Nov 5 '16 at 20:32
  • $\begingroup$ Wow!! I got it!! $\endgroup$
    – bttmbrcelo
    Nov 5 '16 at 20:33
  • $\begingroup$ You are welcome. :) $\endgroup$
    – mesel
    Nov 5 '16 at 20:34
  • $\begingroup$ @bttmbrcelo: If you think that answer is true or usefel, you should upvote or acept as an answer. $\endgroup$
    – mesel
    Nov 5 '16 at 22:31
  • $\begingroup$ oo sorry friend! I forgot to do that!! Thank you once more! $\endgroup$
    – bttmbrcelo
    Nov 5 '16 at 22:54

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