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Abstract Algebra I MATH 4107 Summer 2026

Dr. Hunter Lehmann

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Section Group Actions

Worksheet Part 1: Orbits and Stabilizers

The goal of this problem is to practice understanding the orbits and stabilizers of actions in group theory.

1.

Fill out the table below. \(G\) is a group, \(H\) is a subgroup of \(G\text{,}\) and \(G/H\) is the set of left cosets of \(H\) in \(G\) (not necessarily a group if \(H\) is not normal).
Group Set Action Orbit of \(x\) Stabilizer of \(x\) Fixed Points (\(X_G\))
\(S_{n}\) \(\{1,\ldots, n\}\) \(\sigma\cdot i=\sigma(i)\)
\(G\) \(G\) \(g\cdot x = gx\)
\(G\) \(G\) \(g\cdot x = gxg^{-1}\)
\(H\) \(G\) \(h\cdot x = hx\)
\(G\) \(G/H\) \(g\cdot aH = gaH\)
\(H\) \(G/H\) \(h\cdot gH = hgH\)
\(G\) \(\{H\mid H\leq G\}\) \(g\cdot H = gHg^{-1}\)
Solution.
Here is the table.
Group Set Action Orbit of \(x\) Stabilizer of \(x\) Fixed Points (\(X_G\))
\(S_{n}\) \(\{1,\ldots, n\}\) \(\sigma\cdot i=\sigma(i)\) \(\{1,\ldots, n\}\) \(\{\sigma\in S_n\mid \sigma(i)=i\}\isom S_{n-1}\) None
\(G\) \(G\) \(g\cdot x = gx\) \(G\) \(\{e\}\) None
\(G\) \(G\) \(g\cdot x = gxg^{-1}\) \(\cl(x)=\{gxg^{-1}\mid g\in G\}\) \(C_G(x)\) \(Z(G)\)
\(H\) \(G\) \(h\cdot x = hx\) \(Hx\) \(\{e\}\) None
\(G\) \(G/H\) \(g\cdot aH = gaH\) \(G/H\) \(aHa^{-1}\) None, unless \(H=G\)
\(H\) \(G/H\) \(h\cdot gH = hgH\) \(\{hgH\mid h\in H\}\) \(H\cap gHg^{-1}\) \(N_G(H)/H\)
\(G\) \(\{H\mid H\leq G\}\) \(g\cdot H = gHg^{-1}\) \(\{gHg^{-1}\mid g \in G\}\) \(N_G(H)\) Normal subgroups of \(G\)
The following subgroups may be useful to you.

Definition 123. Centralizer and Normalizer.

If \(S\) is any subset of a group \(G\text{,}\) then the centralizer of \(S\) in \(G\) is the set
\begin{equation*} C_G(S)=\{g\in G \mid gs=sg\text{ for all } s \in S\} \end{equation*}
and the normalizer of \(S\) in \(G\) is the set
\begin{equation*} N_G(S)=\{g\in G \mid gS=Sg \Leftrightarrow gSg^{-1}=S\}\text{.} \end{equation*}
Both the normalizer and centralizer of any set \(S\) are subgroups (not too hard to prove!) and we have \(\{e\}\leq C_G(S) \leq N_G(S)\leq G\) and also \(H \normaleq N_G(H)\) for any subgroup \(H\) of \(G\text{.}\)

Worksheet Part 2: Group Action Proofs

In this part of the activity, you’ll try applying the fundamental theorem of group actions and its consequences to prove some theorems about groups.
In this exercise you will fill in the details of a proof of this theorem using a group action of \(\Z_p\text{,}\) due to James McKay in 1959.

Proof.
We want to find a non-trivial solution to the equation \(g^p=e\) (a solution with \(g\neq \fillinmath{X}\)). To do this, let’s consider the more general equation \(g_1g_2\cdots g_p=e\) in \(p\) variables. Let
\begin{equation*} X=\{(g_1,\dots,g_p)\mid g_i \in G, g_1g_2\dots g_p=e\}\text{.} \end{equation*}
If we choose the elements \(g_1,\dots,g_{p-1}\) freely, then we have \(g_p=\fillinmath{XXXXXXXXXX}\text{.}\) This means the size of \(X\) is \(\fillinmath{XXXXXX}\text{.}\)
Next we show that \(\Z_p\) acts on \(X\) by cyclically permuting the coordinates:
\begin{equation*} k\cdot (g_1,g_2,\dots, g_p)=(g_{k+1},g_{k+2},\dots, g_{k+p}) \end{equation*}
where the indices are taken modulo \(p\text{.}\) This is a group action because:
and if \(g_1g_2\dots g_p=e\) then also \(\fillinmath{XXXXXXXXXXXXXXXXXXXX}\text{.}\)
Now the Fixed Point Congruence shows us that
\begin{equation*} \fillinmath{XXXXX} \equiv |X_{\Z_p}| \bmod{\fillinmath{X}}\text{.} \end{equation*}
A fixed point of this action is exactly a tuple \((g_1,g_2,\dots,g_p)\) where all \(g_i\) are . Since \(p \mid \fillinmath{XXX}\text{,}\) we have also that the right side is . Since there is at least one fixed point, there must be some \(g\neq e\) with .
Here is a filled in version of the proof.

Proof.
We want to find a non-trivial solution to the equation \(g^p=e\) (a solution with \(g\neq e\)). To do this, let’s consider the more general equation \(g_1g_2\cdots g_p=e\) in \(p\) variables. Let
\begin{equation*} X=\{(g_1,\dots,g_p)\mid g_i \in G, g_1g_2\dots g_p=e\}\text{.} \end{equation*}
If we choose the elements \(g_1,\dots,g_{p-1}\) freely, then we have \(g_p=g_{p-1}^{-1}\dots g_2^{-1} g_1^{-1}\text{.}\) This means the size of \(X\) is \(|G|^{p-1}\text{.}\)
Next we show that \(\Z_p\) acts on \(X\) by cyclically permuting the coordinates:
\begin{equation*} k\cdot (g_1,g_2,\dots, g_p)=(g_{k+1},g_{k+2},\dots, g_{k+p}) \end{equation*}
where the indices are taken modulo \(p\text{.}\) This is a group action because:
  • \(\displaystyle 0\cdot(g_1,g_2,\dots,g_p)=(g_1,g_2,\dots,g_p)\)
  • \begin{align*} (m+n)\cdot(g_1,g_2,\dots,g_p) \amp = (g_{m+n+1},g_{m+n+2},\dots,g_{m+n+p})\\ \amp =m\cdot(g_{n+1},g_{n+2},\dots,g_{n+p})\\ \amp =m\cdot(n\cdot (g_1,g_2,\dots,g_p)) \end{align*}
and if \(g_1g_2\dots g_p=e\) then also \(g_{k+1}g_{k+2}\dots g_{k+p}=e\) by substituting \(g_p=(g_1g_2\dots g_{p-1})^{-1}\text{.}\)
Now the Fixed Point Congruence shows us that
\begin{equation*} |X| \equiv |X_{\Z_p}| \bmod{p}\text{.} \end{equation*}
A fixed point of this action is exactly a tuple \((g_1,g_2,\dots,g_p)\) where all \(g_i\) are equal. Since \(p \mid |G|\mid |X|\text{,}\) we have also that the right side is \(0\bmod{p}\text{.}\) Since there is at least one fixed point, there must be some \(g\neq e\) with \(g^p=e\text{.}\)
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