Let \(G\) be a group that acts on \(X\). Define

Definition
Define \(X^G = \{x \in X \ | \ gx = x \text{ for all } g \in G\} \subseteq X \). This set is called the Fixed Point set of the action.


Compare this to the definition for any \(g \in G\), then.

$$ \begin{align*} \text{Fix}(g) = \{x \in X \ | \ gx = x\} \end{align*} $$

So this is the set of elements fixed by \(g\) but what we defined above is the set of elements in \(X\) such that they’re always fixed by any \(g\). This means that we can re-write the definition to

$$ \begin{align*} X^G = \bigcap_{g \in G}\text{Fix}(g) \end{align*} $$

We can also describe this set another way. Recall that an element \(x\) is fixed by every element \(g \in G\) if and only if its orbit contains only the element \(x\) itself. So now we can re-write the definition to be

$$ \begin{align*} X^G = \{x \in X \ | \ O(x) = \{x\} \} \end{align*} $$

Recall now that \(\text{Stab}(x) = \{g \in G \ | \ gx = x\}\). So we can re-write this definition to say

$$ \begin{align*} X^G = \{x \in X \ | \ \text{Stab}(x) = G \} \end{align*} $$



Fixed Point Theorem

We’ll study one theorem about this. But first define

Definition
Let \(p\) be a prime number. A \(p\)-group is a group of order \(p^k\) for some \(k \geq 1\).


For example \(\mathbf{Z}_{p^k}\) is a \(p\)-group. The cyclic group like \(\mathbf{Z}_{p^i} \times \mathbf{Z}_{p^j}\) is another \(p\)-group. Or the dihedral group \(D_{2^k}\) which has order \(2^{k+1}\) so this is a 2-group.

Theorem
Let \(G\) be \(p\)-group which acts on a finite set \(X\). Then \(|X^G| \equiv |X| (\bmod p)\)


Proof

The idea is that since \(G\) acts on \(X\), then it partitions \(X\) into non-empty disjoint orbits. So we can write \(O_1,O_2,...O_r\) for the orbits of the action. Then

$$ \begin{align*} |X| = |O_1| + |O_2| + ... + |O_r| \end{align*} $$

But we know that \(|G|=p^k\) and we also know that any orbit size must divide the group order. Therefore, \(|O_i| \in \{1,p,p^2,...p^k\}\). Break the orbits into two types. Let

$$ \begin{align*} O_1, O_2, ..., O_d \end{align*} $$

be orbits of size 1 and let

$$ \begin{align*} O_{d+1}, O_{d+2}, ..., O_r \end{align*} $$

be orbits of size \(r\). So now

$$ \begin{align*} |X| = (|O_1| + |O_2| + ... + |O_d|) + (|O_{d+1}| + |O_{d+2}| + ... + |O_{r}|) \end{align*} $$

where \(|O_1| + |O_2| + ... + |O_d|=d\) and \(|O_{d+1}| + |O_{d+2}| + ... + |O_{r}|\) is divisible by \(p\) so it’s some multiple \(k\) of \(p\). More precisely, \(d\) is the number of elements that are in orbits of size 1. By definition, this is the fixed set of the action so \(|X^G| = d\). Therefore

$$ \begin{align*} |X| &= d + kp \\ |X| - d &= kp \\ \end{align*} $$

Therefore,

$$ \begin{align*} |X| &\equiv d (\bmod p) \\ |X| &\equiv |X^G| (\bmod p) \end{align*} $$



Application of the Fixed Point Theorem

Here are some application of this theorem

Proposition
Let \(G\) be \(p\)-group. Then the center of the group \(Z(G) = \{ g \in G \ | \ gh = hg \text{ for all } h \in H\}\) is non-trivial so \(Z(G) \neq \{e\}\).


Proof

Let \(G\) acts on \(X = G\) itself by conjugation. So we have \(c: G \rightarrow Sym(X)\). The fixed points of this action are

$$ \begin{align*} X^G &= \{x \in X \ | \ gx = x \text{ for all } g \in G\} \\ &= \{x \in X \ | \ c(g)(x) = x \text{ for all } g \in G\} \quad \text{(the action is the conjugation action)} \\ &= \{x \in X \ | gxg^{-1} = x \text{ for all } g \in G\} \\ &= \{x \in X \ | gx = xg \text{ for all } g \in G\} \end{align*} $$

So in the conjugation action, the fixed point set is the center of the set. By the previous theorem we know that

$$ \begin{align*} |X^G| &\equiv |X| (\bmod p) \\ |X^G| &\equiv p^k (\bmod p) \quad \text{(order of $|G|$ is $p^k$)} \\ |X^G| &\equiv 0 (\bmod p) \quad \text{because ($p^k \equiv 0 (\bmod p)$)}\\ |Z(G)| &\equiv 0 (\bmod p) \quad \text{(we just showed this)}\\ \end{align*} $$

Therefore, \(Z(G) - 0 = pm\) for some \(m \in Z\). This means that \(p \ | \ Z(G)\). But \(Z(G)\) is a subgroup so it includes at least the identity element. So its size is at least 1. Therefore, \(|Z(G)| \geq p \geq 2\). So we must have at least one non-trivial element in the center. \(\ \blacksquare\)

So again, \(p-\)groups will always have a non-trivial center. Next, we have a corollary of this

Proposition
Let \(p\) be a prime number. Then every group of order \(p^2\) is abelian.


Using this, we can now use the elementary divisor theorem to classify these groups. In fact, \(G\) of order (p^2) is isomorphic to either \(\mathbf{Z}_{p^2}\) or \(\mathbf{Z}_p \times \mathbf{Z}_p\).

Proof Let \(|G| = p^2\). By the proposition, \(Z(G)\) is not trivial. But we also know that it is a subgroup. So its order must divide the order of the group. So its order must either be \(p\) or \(p^2\). If the order is \(p^2\), then every element commute with every other element so \(G\) must be abelian. So we only have case which is when \(|Z(G)| = p\). Now, recall that \(Z(G)\) is a normal subgroup in \(G\). Therefore, we can form the quotient group \(G/Z(G)\). The order of this quotient group is \(p^2/p = p\). But we also know that every group of prime order is cyclic so \(G/Z(G)\) is cyclic. By Homework ?, if we have a group \(G\) where its quotient group mod its center is cyclic (\(G/Z(G)\)), then \(G\) is abelian. So \(G\) is abelian in this case too. \(\ \blacksquare\).


Cauchy Theorem

There is one more application of the fixed point theorem.

Theorem
Let \(G\) be a finite group. If \(p\) is a prime number that divides \(|G|\), then \(G\) must have element of order \(p\).


A special case of this is the even-order theorem. As a reminder, let’s revisit this proof in terms of group actions before proving the actual theorem.

Proof (Even Order Theorem)
Let \(|G| = n\) where \(n\) is even. Let \(C = \langle \varphi \rangle = \{e, \varphi\}\) be a cyclic group of order 2. Let \(C\) act on the set \(X = C\) which is the group itself. We know the identity elements acts as the identity function on \(X\). So we only have to define the action for \(\varphi\). So let’s define what \(\varphi(g)\) is for every element of the group. Let \(\varphi(g)\) be

$$ \begin{align*} \varphi(g) = g^{-1} \quad \text{for } g \in G \end{align*} $$

This is in fact is a bijection. Note here, \(\varphi \circ \varphi = id\) so composing the action \(\varphi\) with itself gives us back the identity function. Now, \(C\) is a 2-group since its of order 2. So we can apply the fixed point theorem which states that

$$ \begin{align*} |X^C| &\equiv |X| (\bmod 2) \\ |X^C| &\equiv 2 (\bmod 2) \quad \text{(We know $|X| = 2$)}\\ |X^C| &\equiv 0 (\bmod 2) \end{align*} $$

So \(|X^C|\) must be even. \(X^C\) is the set of elements that are fixed by the action \(\varphi\) so

$$ \begin{align*} X^C &= \{x \in X \ | \ \varphi(x) = x\} \\ &= \{x \in X \ | \ x^{-1} = x\} \\ &= \{x \in X \ | \ x^2 = e\}. \end{align*} $$

So this group has an even number of elements and must at least include the identity element. Therefore, it must have at least one more non-trivial element of order 2. \(\ \blacksquare\)


Proof (Cauchy’s Theorem)
Let \(G\) be a group of order \(n\). Suppose \(p\) is a prime number such that \(p \ | \ n\). Let \(C = \langle \varphi \rangle\) be a cyclic group of order \(p\). Let \(X\) be the set

$$ \begin{align*} X = \{(a_1,...,a_p) \in G^p \ | \ a_1,...a_p \in G, a_1a_2...a_p = e\} \end{align*} $$

So it’s the set of all \(p\) tuples such that when we multiply any tuple’s elements, we get the identity. But we can re-write this as

$$ \begin{align*} a_p = (a_1,a_2...a_{p-1})^{-1} \end{align*} $$

So the last element \(a_p\) is the inverse of the previous elements all multiplied. So for any \(g\) to be in \(G^p\), we can pick \(p-1\) elements from \(G\) and then form the last element by taking their product and taking the inverse of that product.

What is \(|X|\)? we have \(n\) choices for the first \(p-1\) elements but only 1 choice for the last element. This implies that \(|X| = n^{p-1}\). Moreover, by assumption we know that \(p\) divides \(|G|=n\). So \(n = pk\) for some \(k\). So we can write \(|X| = (pk)^{p-1}\). But \(p\) is prime so it’s at least 2. Therefore, \(p\) must divide \(|X|\) as well.

Now, let \(C\) act on \(X\). Define

$$ \begin{align*} \varphi \cdot (a_1,a_2,...,a_{p-1},a_p) = (a_p,a_1,...,a_{p-1}) \end{align*} $$

So the action permutes the elements cyclicly. We need to verify that the product is in \(X\). This means if we multiply the first \(p-1\) elements and take their inverse, we should get the last element. To show this notice that \((a_1,a_2,...,a_{p-1},a_p)\) is in \(X\) by assumption so we know that the product of the elements is \(e\). Now conjugate this product by \(a^{p}\) to see that

$$ \begin{align*} (a_1a_2...a_{p-1}a_p) &= e \\ a_p(a_1a_2...a_{p-1}a_p)a_p^{-1} &= a_pea_p^{-1} \\ a_pa_1a_2...a_{p-1} &= e. \end{align*} $$

So we can see that \(a_pa_1a_2...a_{p-1} \in X\) which is what we wanted to show. Additionally, if we apply \(\varphi\) \(p\) times, we will see that \(\varphi \circ \varphi \circ ... \varphi = id\) it will take us to the identity function or action. So this action or permutation has order \(p\).

So now we have \(|G| = n\) and \(|X| = n^{p-1}\). We know \(p\) divides both. We can apply the fixed point theorem but what is \(X^C\)? By definition, it’s the set of elements fixed by any \(g \in \langle \varphi \rangle\). But since \(\langle \varphi \rangle\) is cyclic, then if an element gets fixed by \(\varphi\), it get fixed by any power of \(\varphi\). Therefore

$$ \begin{align*} X^C &= \{x \in X \ | \ \varphi(x) = x\} \\ &= \{(a_1,...,a_p) \in G \ | \ a_1...a_p = e \text{ and } \varphi \cdot (a_1,...,a_p) = (a_1,...,a_p)\} \\ &= \{(a_1,...,a_p) \in G \ | \ a_1...a_p = e \text{ and } (a_p,a_1...,a_{p-1}) = (a_1,a_2...,a_p)\}. \end{align*} $$

This last condition says that \(a_p=a_1\), \(a_1=a_2\), … \(a_{p-1}=a_p\). This means that all the elements are the same. So we can write \(X^C\) as

$$ \begin{align*} X^C &= \{ (a,a...,a) \ | \ a \in G, aa...a = a^p = e\}. \end{align*} $$

Therefore, the size of this set, is the number of elements in \(G\) which have order \(p\). So

$$ \begin{align*} |X^C| &= |\{ a \in G \ | \ a^p = e\}|. \end{align*} $$

We know \(e \in X^C\) since e\(e^p = e\). Moreover, by the Fixed Point Theorem,

$$ \begin{align*} |X^C| &\equiv |X| (\bmod p) \\ |X^C| &\equiv 0 (\bmod p) \quad \text{(because $p \ | \ |X|$)} \end{align*} $$

So \(|X^C|\) must be divisible by \(p\) and since \(|X^C| \geq 1\), then \(|X^C| \geq p\). But this means that \(G\) has at least one non-trivial element of order \(p\). \(\ \blacksquare\)


Classification of Groups of Order 6

We’ve seen before \(\mathbf{Z}_6 \cong \mathbf{Z}_2 \times \mathbf{Z}_3\) and we’ve also seen \(D_3 \cong S_3\).

Suppose \(|G| = 6 = 2(3)\). These are prime factors, so we can use Cauchy’s Theorem twice to conclude that we must have an element of order 2 and another element of order 3. So let

$$ \begin{align*} A &= \langle a \rangle \text{ where $|A| = 2$ } \\ N &= \langle 3 \rangle \text{ where $|N| = 3$} \end{align*} $$

This implies that

$$ \begin{align*} [G:N] = \frac{|G|}{|N|} = \frac{6}{3} = 2. \end{align*} $$

But we’ve seen in one of the homeworks, that this implies that \(N\) is normal. We also know the following facts

  • \(N \cap A = \{e\}\) since elements of \(N\) have order 1 or 3 and elements of \(A\) have orders 1 and 2
  • \(NA\) is a subgroup of \(G\) because one of the groups is normal by Corollary (26.4).
  • By the Diamond Isomorphism Theorem, \(A/A \cap N \cong NA/N\). But since \(A \cap N = \{e\}\). Then \(A \cong NA/N\). This means that \(|NA|/|N| = |A|\). So \(|NA| = |A||N| = 6\)
  • Since \(NA\) is a subgroup of size 6, then it's \(G\) so \(NA = G\)

So we know 4 important things

  • \(A\) is a subgroup of \(G\).
  • \(N\) is a normal subgroup of \(G\).
  • \(G = NA\).
  • \(A \cap N = \{e\}\).

These are the 4 conditions so we can apply the the recognization theorem to conclude that there exists a homomorphism \(\gamma: A \rightarrow \text{Aut}(N)\) such that there is an isomorphism of groups

$$ \begin{align*} N \rtimes_{\gamma} A \cong G \end{align*} $$

\(A\) is of order 2 so it must be isomorphic to \(\mathbf{Z}_2\). \(N\) is of order 3 so it’s isomorphic to \(\mathbf{Z}_3\). We know that \(\text{Aut}(\mathbf{Z}_3) \cong \Phi(3)\) But \(\Phi(3)\) is of order 2 so it’s isomorphic to \(\mathbf{Z}_2\). So how many homomorphisms can we have from \(A\) to \(\text{Aut}(N)\) if both groups are cyclic of order 2? There are only two choices.

  • The trivial homomorphism gives us the product group \(\mathbf{Z}_2 \times \mathbf{Z}_3\)
  • The non-trivial homomorphism gives us the dihedral group \(D_3\).



Study Notes on \(D_3\)

How does the non-trivial homomorphism gives us the dihedral group? how does this happen? We have

  • \(N = \langle n \rangle \cong \mathbf{Z}_3\) where \(n^3 = e\).
  • \(A = \langle a \rangle \cong \mathbf{Z}_2\) where \(a^2 = e\)

Define

$$ \begin{align*} \gamma: A &\rightarrow \text{Aut}(\mathbf{Z}_3) \\ \gamma(a) &= \alpha \text{ where } \alpha(n) = n^{-1} \\ \gamma(a)(n) &= n^{-1} \end{align*} $$

So the element \(a \in A\) acts on \(n \in N\) by inverting it. Observe that \(a^2 = e\) and \(\alpha^2 = (n^{-1})^{-1} = \text{id}\) so \(\gamma(\alpha^2) = \gamma(\alpha)^2\). So now multiplication in semi-direct groups is defined as

$$ \begin{align*} (n^i, a^j)(n^k, a^l) = (n^i \cdot \gamma_{a^j}(n^k), a^{j}a^{j}) \end{align*} $$

\(a\) has order 2 so \(\gamma_{a^j}(n^k)\) is defined as

  • When \(j = 0\), then \(a^0 = 0\), then \(\gamma_{e} = id\).
  • When \(j = 1\), then \(a^1 = a\), then \(\gamma_{a} = \alpha\) where \(\alpha(n) = n^{-k}\).

So now if we apply the semidirect product multiplication, using the homomorphism we defined, we will see that for any \(a\), that we get the relationship \(ana^{-1} = n^{-1}\).


Classification of Groups of Order \(2p\)

where \(p\) is an odd prime. We know two groups \(\mathbf{Z}_{2p} =\mathbf{Z}_{2} \times \mathbf{Z}_{p}\) and \(D_p\).

Let \(|G| = 2p\). By Cauchy, \(2\) divides \(|G|\). Therefore, we have an element of order \(2\). From this, we get \(A = \langle a \rangle\) where \(|A| = 2\). We also have an element of order \(p\). From this we get \(N = \langle N \rangle\). Again, we will see that

$$ \begin{align*} [G:N] = \frac{|G|}{|N|} = \frac{2p}{p} = 2. \end{align*} $$

Therefore \(N\) is normal. With a similar argument to the previous example. We will see that

  • \(A\) is a subgroup of \(G\).
  • \(N\) is a normal subgroup of \(G\).
  • \(G = NA\).
  • \(A \cap N = \{e\}\).

So we can use the recognization theorem again to conclude that there exists a homomorphism \(\gamma: A \rightarrow \text{Aut}(N)\) such that there is an isomorphism of groups

$$ \begin{align*} N \rtimes_{\gamma} A \cong G \end{align*} $$

\(A\) is of order 2 so it must be isomorphic to \(\mathbf{Z}_2\). \(N\) is of order \(p\) so it’s isomorphic to \(\mathbf{Z}_p\) (cyclic). We know that \(\text{Aut}(\mathbf{Z}_p) \cong \Phi(p)\) where \(\Phi(p)\) is of order \(p-1\). So how many homomorphisms can we have from \(A\) to \(\text{Aut}(N)\).

  • The trivial homomorphism gives us the product group \(\mathbf{Z}_2 \times \mathbf{Z}_p\)
  • The non-trivial homomorphism gives us the dihedral group \(D_p\).




Classification of Groups of Order \(pq\)

\(p\) and \(q\) are distinct primes where \(p > q\). So again we have \(|G|=pq\). By Cauchy, we have two cyclic subgroups such that

$$ \begin{align*} A &= \langle a \rangle \text{ where $|A| = q$ } \\ N &= \langle b \rangle \text{ where $|N| = p$} \end{align*} $$

This implies that

$$ \begin{align*} [G:N] = \frac{|G|}{|N|} = q. \end{align*} $$

\(q\) is the smallest prime dividing the order of \(|G|\). By some homework assignment that said that if we have a subgroup of order “the smallest prime dividing the order”, then this subgroup is normal. So now again,

  • \(A\) is a subgroup of \(G\).
  • \(N\) is a normal subgroup of \(G\).
  • \(G = NA\).
  • \(A \cap N = \{e\}\) because \(p \neq q\).

So we can use the recognization theorem AGAIN to conclude that there exists a homomorphism \(\gamma: A \rightarrow \text{Aut}(N)\) such that there is an isomorphism of groups

$$ \begin{align*} N \rtimes_{\gamma} A \cong G \end{align*} $$

Where

$$ \begin{align*} \gamma: A &\rightarrow \text{Aut}(N) \\ \gamma: \mathbf{Z}_p &\rightarrow \Phi(p) \end{align*} $$

There are two cases:

  • \(q \ \not\mid \ (p-1)\): In this case, the generator of the group \(A\)'s \(q\)th power has to go to the identity because \(q\) doesn't divide \(p - 1\) which is the order of \(\Phi(p)\). The only possible \(\gamma\) is \(\gamma(a) = e\) so send everything to the identity and we get the direct product \(\mathbf{Z}_p \times \mathbf{Z}_q\).
  • \(q \ | \ (p-1)\): So we get the non-trivial homomorphism \(\gamma\). Because \(q \ (p-1)\) which is the order of \(\Phi(p-1)\), then by Cauchy there exists an element of order \(q\) in \(\Phi(p-1)\). So \(\mathbf{Z}_{pq}\) and another non abelian group.




References