Let $N$ be a normal subgroup of a finite group $G$. Let $S$ belonging to $G$ be a conjugacy class of elements in $G$, and assume that $S$ belongs to $N$. Prove that $S$ is a union of $n$ conjugacy classes in $N$, all having the same cardinality, where $n$ equals the index $[G : NCent(x)]$ of the group generated by $N$ and the centralizer $Cent(x)$ in $G$ of and element $x$ belonging to $S$.

  • 5
    $\begingroup$ Homework question about group? $\endgroup$ – Aryabhata Sep 28 '10 at 5:28
  • $\begingroup$ i am learning abstract algebra right now. it is just an exercise (not homework) i am trying to do. i have been thinking about it for several days, but cannot find a way. $\endgroup$ – Yuan Sep 28 '10 at 10:30
  • 2
    $\begingroup$ I hope there is a more useful title for your question! $\endgroup$ – Mariano Suárez-Álvarez Sep 28 '10 at 11:49
  • 1
    $\begingroup$ When you say "belongs", you mean "contained", right? "Belongs" usually is interprete to mean "is an element of", but that cannot be what you mean here. $\endgroup$ – Arturo Magidin Sep 28 '10 at 14:11

The fact that $S$ is a union of conjugacy classes of $N$ should be immediate; also, the sizes are the same: let $x\in S$, and let $U$ be the $N$-conjugacy class of $x$. If $T$ is any other $N$-conjugacy class contained in $S$, then let $y\in T$. There exists $g\in G$ such that $gxg^{-1}=y$. Then $|T|=|gUg^{-1}|=|U|$.

So the real meat of the problem is showing that the number of conjugacy classes is the index of $NC_G(x)$ in $G$.

Whenever you see a problem in which you need to show that the number of something or other in a group equals the index of a subgroup $H$ in $G$, this should suggest finding a map from $G$ to the set you want, and then showing that the images of two elements coincide if and only if they lie in the same coset of $H$ in $G$. This will then give you that the number of distinct images is the index of $H$ in $G$.

So here what you want to do is find a map from $G$ to the conjugacy classes of $N$ contained in $S$, and arrange matters so that two elements of $G$ map to the same thing if and only if they lie in the same coset of $NC_G(x)$.

So... how to map it? The obvious thing to do is to take a conjugacy class of $N$ contains in $S$, represented by some $x\in S$, and map it to the conjugacy class of $gxg^{-1}$ (which lies in $N$ because $N$ is assumed to be normal). If $x\sim_N y$ (that is, $x$ is conjugate to $y$ in $N$) then $y=nxn^{-1}$, so $gyg^{-1} = gnxn^{-1}g^{-1} = (gng^{-1})x(gng^{-1})$, and since $gng^{-1}\in N$, then $gyg^{-1}\sim_N gxg^{-1}$. So this map is well-defined on $N$-conjugacy classes. Since $S$ is a conjugacy class, $gxg^{-1}\in S$ as well, so the image of the $N$-conjugacy class of $x$ is also an $N$-conjugacy class contained in $S$. That is: conjugation maps $N$-conjugacy classes contained in $S$ to $N$-conjugacy classes contained in $S$. And note that this map is onto: if you pick any $x\in S$, then every element of $S$ is of the form $gxg^{-1}$ for some $g\in G$, so you can pick your favorite $x\in S$, and use it as a basis: you will "hit" every $N$-conjugacy class that makes up $S$ this way.

Now, when will $g$ and $h$ do the exact same thing to every $N$-conjugacy class contained in $N$? Suppose that $gxg^{-1}\sim_N hxh^{-1}$ for some $x\in S$. Then there exists $n\in N$ such that $ngxg^{-1}n^{-1} = hxh^{-1}$, hence $h^{-1}ngxg^{-1}n^{-1}h = x$; that is, there exists $n\in N$ such that $h^{-1}ng\in C_G(x)$ (we're getting somewhere... we've gotten to the centralizer of $x$ in $G$). This means that $ng C_G(x)=hC_G(x)$. Not quite what we want: that $n$ is "on the wrong side" so to speak. Ah, but $N$ is normal. So $ng = (gg^{-1})ng = g(g^{-1}ng) = gn'$ for some $n'\in N$, so we have $gn'C_G(x)=hC_G(x)$. Thus, of $gxg^{-1}\sim_N hxh^{-1}$, then there exists $n'\in N$ such that $gn'C_G(x)=hC_G(x)$. And this is equivalent to $gNC_G(x) = hNC_g(x)$. (Almost there!) Conversely, suppose that $gNC_G(x) = hNC_g(x)$; then there exists $n\in N$ and $c\in C_G(x)$ such that $h = gnc$. Then $hxh^{-1} = (gnc)x(gnc)^{-1} = gn(cxc^{-1})n^{-1}g^{-1} = gnxn^{-1}g^{-1}$ (since $cx=xc$); since $n$ is normal, $gn = gn(g^{-1}g) = (gng^{-1})g = n'g$ for some $n'\in N$, so $hxh^{-1} = gnxn^{-1}g^{-1} = n'gxg^{-1}n'^{-1}$, hence $hxh^{-1}\sim_N gxg^{-1}$. That is: $g$ and $h$ map the $N$-conjugacy class of $x$ to the same $N$-conjugacy class if and only if $g$ and $h$ are in the same left coset of $NC_G(x)$ in $G$.

Now, since every element of $S$ is $G$-conjugate to $x$, this is enough: we can get to any $N$-conjugacy class contained in $N$ by conjugating $x$ by some element of $G$. And we have just seen that two elements of $G$ have the same image if and only if they are in the same left cosets of $NC_G(x)$. So the number of distinct images, that is, the number of $N$-conjugacy classes that make up $S$ (and here we can use "number" to mean cardinality, even in the infinte case!), is exactly the same as the number of left cosets of $NC_G(x)$ in $G$. That is, the number of conjugacy classes is $[G:NC_G(x)]$, as claimed.


for $x \in S$ define a function from $G/ NCent(x)$ to conjugacy classes in N by $gNCent(x) \mapsto [gxg^{-1}]$. This is well defined because changing by an element of Cent(x) does nothing to x, and changing by an element of N changes $gxg^{-1}$ to another element in the same conjugacy class of N.

It is easy to show that this is onto the conjugacy classes in S. also this is injective, because if $[hxh^{-1}]=[gxg^{-1}]$ then there is an $m\in N$ such that $(g^{-1}mh)x=x(g^{-1}mh)$, so you have that $g^{-1}h (h^{-1}mh) \in Cent(x)$.

let $m_i\in N$ be the elements such that ${ m_i x m_i ^{-1} }$ are all the elements in the conjugacy class of x in N. From this we get that $(g_i m_i g_i^{-1}) (g_i x g_i^{-1}) (g_i m_i g_i^{-1})^{-1} $ are different elements in the conjugacy class of $(g_i x g_i^{-1})$ so $ | [x] | \leq | [gxg^{-1}] | $, but there is nothing special about taking x from S so we have the opposite inequality as well.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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