Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to understand these two derivations in class.

Here $g \in G$ and $x \in S$.

$\displaystyle\sum_{x \in S} |G_x| = \sum_{\text{orbits} \hspace{ 1mm} Gx} \hspace{ 1mm} \sum_{y \in Gx} |G_y| = \sum_{\text{orbits} \hspace{ 1mm} Gx} |Gx| |G_x| = \sum_{\text{orbits} \hspace{1mm} Gx} |G| = |S/G||G| $

What I mainly don't understand is the double summation as in how: $\displaystyle\sum_{\text{orbits} \hspace{ 1mm} Gx} \hspace{ 1mm} \sum_{y \in Gx} |G_y| = \sum_{\text{orbits} \hspace{1 mm} Gx} |Gx| |G_x|$

My other question is how do we know that $\displaystyle\sum_{y \in Gx} 1 = |Gx|$

share|cite|improve this question
up vote 4 down vote accepted

To take the second question first: the sum $\sum\limits_{y\in Gx}1$ has one term for every $y\in Gx$, and each of those terms is $1$, so it’s just adding $|Gx|$ copies of the number $1$; the result, of course, is $|Gx|$. If you want to make that even more explicit, let $m=|Gx|$ and enumerate the elements of $Gx$ as $y_1,\dots,y_m$. For each $y\in Gx$ let $f(y)=1$. Then $$\sum_{y\in Gx}1=\sum_{y\in Gx}f(y)=\sum_{k=1}^mf(y_k)=\sum_{k=1}^m 1=m\cdot1=m=|Gx|\;.$$

Now let’s look at the first question. You’re starting with $$\sum_{\text{orbits} \hspace{ 1mm} Gx} \hspace{ 1mm} \sum_{y \in Gx} |G_y|\;.\tag{1}$$ I think that this might be clearer if we gave a name to the set of orbits: let $\Omega$ be the set of orbits. Then $(1)$ can be rewritten as $$\sum_{\omega\in\Omega}~\sum_{y\in\omega}|G_y|\;.\tag{2}$$ The next thing to realize is that if $x$ and $y$ are in the same orbit, then $|G_x|=|G_y|$.

Proof: If $x$ and $y$ are in the same orbit, then $y=gx$ for some $g\in G$. Then $h\in G_y$ iff $hy=y$ iff $hgx=gx$ iff $g^{-1}hgx=x$ iff $g^{-1}hg\in G_x$, and the map $h\mapsto g^{-1}hg$ is a bijection. $\dashv$

Thus, for any orbit $\omega\in\Omega$ there is a number $n(\omega)$ such that $|G_x|=n(\omega)$ for each $x\in\omega$. Thus, $(2)$ can be rewritten as $$\sum_{\omega\in\Omega}~\sum_{y\in\omega}n(\omega)=\sum_{\omega\in\Omega}|\omega|\,n(\omega)\;,\tag{3}$$ since the inner sum on the lefthand side is just adding up $|\omega|$ copies of the number $n(\omega)$.

We could now pick out a particular element $x_\omega$ of each orbit $\omega$ use these elements to identify the orbits. If we do this, $(3)$ becomes


which is just a slightly more careful way of writing

$$\sum_{\text{orbits}} |Gx| |G_x|\;.$$

share|cite|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.