# a question about group (decomposition of conjugacy classes in normal subgroups)

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 $S$.

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Homework question about group? – Aryabhata Sep 28 '10 at 5:28
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. – Yuan Sep 28 '10 at 10:30
I hope there is a more useful title for your question! – Mariano Suárez-Alvarez Sep 28 '10 at 11:49
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. – 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.

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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.

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