1.10. Conjugation, The Class Equation, and Cauchy's Theorem.

\textcolor{blue}{We now touch on a very deep example of a group action, known as conjugation.} Let $G$ act on itself "by conjugation", which we define as follows. Let $g, h \in G$. Then

\[ g * h = ghg^{-1} \]

is the group action of conjugation. Let's show that this is a group action.

Composition. Let $g_1, g_2, h \in G$. Then observe that

\[\begin{align*} g_1 * (g_2 * h) = g_1 * (g_2hg_2^{-1}) &= g_1g_2 h g_2^{-1}g_1^{-1} \\ & = (g_1g_2) h (g_1g_2)^{-1} \\ & = (g_1g_2) * h \end{align*}\]

so that the first axiom of a group action is satisfied. * Identity. Observe also that for $e \in G$, the identity of $G$,

\[ e * h = ehe^{-1} = h. \]

Therefore this is a group action.

We'll now show that this group action is very special and important. Conjugation itself is important in math. In Linear Algebra, two matrices which are similar (i.e., $A$ is similar to $B$ if there exists $P$ such that $A = P^{-1}BP$) have the same rank, determinants, trace, eigenvalues, and much more. Basically, they represent the same linear transformation, just in different bases. To learn more about conjugation, we make a few definitions with this group action.

Let $G$ be a group, and let $G$ act on itself by conjugation. For any $h \in G$, the orbit of this group action

\[\begin{align*} Gh & = \{g * h \mid g \in G\} \\ & = \{ghg^{-1} \mid g \in G\} \end{align*}\]

is known as a conjugacy class of $G$.

\textcolor{purple}{Previously we discussed how orbits of a group action partition the set $X$ which is being acted on. Since $G$ acts on itself in this example, we see that the conjugacy classes form a partition of $G$!} \ \ Remark. Recall the definition of a centralizer $G$ for a set $A \subset G$:

\[\begin{align*} C_G(A) & = \{g \in G \mid gs = sg \text{ for all } s \in S\}\\ & = \{g \in G \mid gsg^{-1} = s \text{ for all } s \in S\}. \end{align*}\]

Therefore for a single point $x \in G$, $C_G(x) = \{g \in G \mid gxg^{-1} = x \} = \{g \in G \mid g * x = x\}$, where in the last equation we are speaking in terms of group actions. But note that this last set is exactly the stabilizer of $G$ under this group action. \textcolor{NavyBlue{Therefore, $C_G(x) = G_x$ for any $x \in G$ under this group action.}}

Furthermore, let $x \in Z(G) = \{z \in G \mid z = gzg^{-1} \text{ for all } g \in G\},$ the center of $G$. Then we see that $Gx = \{gxg^{-1} \mid g \in G\} = \{x\}$. \textcolor{NavyBlue{So for any $x \in Z(G)$, the orbit is of size one. The sole element it contains is just $x$.}} (We can go even further: the conjugacy classes of an abelian group are all of size one.) \ \ Let's put all of these results together. In general, if $G$ acts on itself via conjugation, then we know its orbits, or conjugacy classes, partition $G$. Moreover, let $R \subset X$ be a set of orbit representatives (or conjugacy class representatives, if you like). Then

\[ |G| = \sum_{x \in R}|Gx| \]

Recall that $|Gx| = 1$ if $x \in Z(G)$. Thus we can write this further as

\[\begin{align*} |G| = \sum_{x \in Z(G)}|Gx| + \sum_{x \in R\setminus Z(G)} |Gx| & = \sum_{x \in Z(G)}1 + \sum_{x \in R\setminus Z(G)} |Gx|\\ & = |Z(G)| + \sum_{x \in R\setminus Z(G)} |Gx| \end{align*}\]

By the Orbit-Stabilizer theorem, we can write $|Gx| = |G|/|G_x|$. Substituting this in, we get

\[\begin{align*} |G| &= |Z(G)| + \sum_{x \in R\setminus Z(G)} |G|/|G_x| \end{align*}\]

and since $C_G(x) = G_x$,

\[\begin{align*} |G| = |Z(G)| + \sum_{x \in R\setminus Z(G)} |G|/|C_G(x)| \end{align*}\]

which is known as the class equation. This equation is pretty badass, as it gives us a way to understand the cardinality of a group. This equation is also useful in proofs, as we shall see in the following examples. First, we begin with a lemma.

Let $G$ be a group. Then $C_G(x) = G$ if and only if $x \in Z(G)$.

Suppose $C_G(x) = G$. Then for all $g \in G$, $gx = xg$. However, $Z(G)$ is the set of all $G$ which commutes with every member of $G$, so $x \in Z(G)$. Now suppose $x \in Z(G)$. Then $gx = xg$ for all $g \in G$. Therefore, $C_G(x) = \{g \in G \mid gx = xg\} = G$.

Let $G$ be a group such that $|G| = p^n$ for some prime $p$ and $n \in \mathbb{N}$. Then $|Z(G)| > 1$. That is, $|Z(G)| \in \{p, p^2, \dots, p^{n}\}$. \textcolor{Purple}{Equivalently, this theorem says that $Z(G)$ is nontrivial. Moreover, this implies that **there exists non identity elements of $\mathbf{G**$ which commute with every element of $\mathbf{G}$.}}

First observe that $Z(G)$ is a subgroup of $G$. Therefore, by Lagrange's Theorem, we know that $|Z(G)|$ divides $G$. Thus $|Z(G)| \in \{1, p, p^2, \dots, p^{n}\}$. Our goal is to show that $|Z(G)|$ cannot equal 1.

\textcolor{NavyBlue}{For the sake of contradicton, suppose $|Z(G)| = 1$}. Then by the previous lemma, we see that there is no nontrivial element $g$ of $G$ such $C_G(g) = G$.

Let $R$ be the set of conjugacy class representatives. Then $|G|/|C_G(r)| \in \{p, p^2, \cdots, p^n\}$ for $r \in R\setminus Z(G)$ (since $|Z(G)| = 1$, $R\setminus Z(G)$ simply removes $e$, the identtiy, from $R$).

\textcolor{red!40!purple!100}{Why can't $|G|/|C_G(r)| = 1$ for any $r \in R\setminus Z(G)$? Well, because for such an $r$, $r \not\in Z(G)$. Therefore $C_G(r) \ne G$, so $|G|/|C_G(r)| \ne 1$.}

Now by the class equation, we see that

\[ \underbrace{ \hspace{.2cm}|G|\hspace{.2cm} }_{\text{divisible by } p} \hspace{-.5cm} = |Z(G)| + \overbrace{\sum_{r \in R\setminus Z(G)} |G|/|C_G(r)|}^{\text{divisible by } p} \]

since $|G|/|C_G(r)| \in \{p, p^2, \dots, p^n\}$ for all $r \in R\setminus Z(G)$. \textcolor{NavyBlue}{Therefore we see that $|Z(G)|$ must be divisible by $p$. But this is a contradiction since we said $|Z(G)| = 1$}. Therefore, we see that $|Z(G)| \in \{p, p^2, \dots, p^n\}$.

The above theorem can be used to prove the next theorem, whose signifiance demonstrates the power of the class equation. The theorem below is generally proved by proving the above theorem first in the special case for when $|G| = p^2$. But it will be helpful to other proofs later on to consider the more general case as we presented it above.

Let $G$ be a group, and suppose $|G| = p^2$ where $p \ge 2$ is prime. Then $G$ is abelian.

\textcolor{green!50!black}{By the previous theorem, we see that $|Z(G)| \in \{p, p^2\}$.)} We'll proceed by considering two cases.

$\mathbf{|Z(G)| = p^2}$. In this case $|G| = |Z(G)|$. Since we also have that $Z(G)$ is a subgroup of $G$, we can conclude that $G = Z(G)$. Therefore, $G$ is abelian.
$\mathbf{|Z(G)| = p}$. Recall that $Z(G) \normal G$ from Proposition \ref{normal_center}. Therefore, we can speak of the quotient group $G/Z(G)$, which has size $|G|/|Z(G)| = p^2/p = p$. By the corollary to Lagrange's Theorem, this implies that $G/Z(G)$ is cyclic, since it has prime order. Thus there exists a $g \in G$ such that we can represent $G/Z(G)$ as

\[ G/Z(G) = \{Z(G), Z(G)g, Z(G)g^2, \dots, Z(G)g^{p-1}\}. \]

As we already know, cosets partition $G$. Therefore, let $a, b \in G$, and suppose $a \in Z(G)g^i$ and $b \in Z(G)g^j$. Then there exist $x, y \in Z(G)$ such that $a = xg^i$ and $b = yg^j$. Thus observe that

\[\begin{align*} ab = xg^i yg^j = xyg^ig^j = xyg^{i+j} = xyg^jg^i = yg^jxg^i = ba \end{align*}\]

where we used the commutavity of $x,y$ since $x, y \in Z(G)$. Since $a, b$ were arbitrary members of $G$, this proves that $G$ is abelian.

Thus we see that the class equation is useful in proving more general facts about group theory. The class equation can also be used to prove the following important theorem in group theory, known as Cauchy's Theorem.

[ (Cauchy's Theorem)] Let $G$ be a finite group and $p \ge 2$ be a prime. If $p$ divides the order of $G$, then $G$ has an element of order $p$.

\noindent\textcolor{NavyBlue}{So consider a group $G$ with order $n$, and suppose

\[ n = p_1^{i_1}\cdot p_2^{i_2} \cdots p_n^{i_n} \]

is its prime factorization. Then there exist elements $g_1, g_2, \dots, g_n$ such that $|g_i| = p_i$ for $i = 1, 2, \dots, n$.}

Another way to visualize this as follows. Consider a group $G$ consisting of 10 elements.

\[ \{e, \hspace{.1cm}g_1,\hspace{.1cm} g_2,\hspace{.1cm} g_3,\hspace{.1cm} g_4,\hspace{.1cm} g_5,\hspace{.1cm} g_6,\hspace{.1cm} g_7,\hspace{.1cm} g_8,\hspace{.1cm} g_{9}\} \]

By Cauchy's theorem, there exists elements of order $2$ and $5$. So suppose $g_1$ and $g_2$ are such elements, i.e., $g_1^2 = 3$ and $g_2^5 = e$. Then we can really rewrite this as

\[ \{e,\hspace{.1cm} \textcolor{red}{g_1},\hspace{.1cm} \textcolor{blue}{g_2},\hspace{.1cm} \textcolor{blue}{g_2^2},\hspace{.1cm} \textcolor{blue}{g_2^3},\hspace{.1cm} \textcolor{blue}{g_2^4},\hspace{.1cm} g_6,\hspace{.1cm} g_7,\hspace{.1cm} g_8,\hspace{.1cm} g_9\}. \]

However, we know $\textcolor{red}{g_1}\textcolor{blue}{g_2}, \textcolor{red}{g_1}\textcolor{blue}{g_2^2}, \textcolor{red}{g_1}\textcolor{blue}{g_2^3}$ and $\textcolor{red}{g_1}\textcolor{blue}{g_2^4}$ are all in $G$. Thus we can really write this as

\[ \{e, \hspace{.1cm} \textcolor{red}{g_1},\hspace{.1cm} \textcolor{blue}{g_2},\hspace{.1cm} \textcolor{blue}{g_2^2},\hspace{.1cm} \textcolor{blue}{g_2^3},\hspace{.1cm} \textcolor{blue}{g_2^4}, \hspace{.1cm}\textcolor{red}{g_1}\textcolor{blue}{g_2},\hspace{.1cm} \textcolor{red}{g_1}\textcolor{blue}{g_2^2},\hspace{.1cm} \textcolor{red}{g_1}\textcolor{blue}{g_2^3},\hspace{.1cm} \textcolor{red}{g_1}\textcolor{blue}{g_2^4}\}. \]

Thus we can understand the structure of every single group of order 10. But this can be done for all finite groups!

\textcolor{Plum}{In this proof, we'll prove this in a very clevery way by letting a subgroup of a permutation group act on a special set $X$ (both of which we will define). This will then prove the existence of elements of order $p$.}

Let $p$ be a prime which divides $|G|$. Define $H$ to be the cyclic subgroup of $S_p$ generated by $(1\hspace{.1cm}2\hspace{.1cm}\cdots\hspace{.1cm}p)$.

We can picture $H$ as the group

\[ \{(1\hspace{.1cm}2\hspace{.1cm}\cdots\hspace{.1cm}p), (2\hspace{.1cm}3\hspace{.1cm}\cdots\hspace{.1cm}p, \hspace{.1cm}1), \cdots, (p\hspace{.1cm}1\hspace{.1cm}\cdots\hspace{.1cm}p-1)\}. \]

Now let $H$ act on the set $X$ defined as

\[ X = \{ (g_1, g_2, \dots, g_p) \mid g_1, g_2, \dots, g_p \in G \text{ and } g_1g_2\cdots g_p = e \} \]

where the $\sigma \in H$ acts on $g \in X$ as

\[ \sigma \cdot (g_1, g_2, \dots, g_p) = (g_{\sigma(1)}, g_{\sigma(2)}, \dots, g_{\sigma(p)}). \]

This $H$ takes a $p$-tuple in $X$ and permutates the elements. Since $H$ is generated by $(1\hspace{.1cm}2\hspace{.1cm}\cdots\hspace{.1cm}p)$, it "pushes" the elements $g_i$ in the tuple over to the right, and the elements that are pushed out of the right end of the tuple are pushed back in on the left side.

\textcolor{NavyBlue}{First we'll show that this is a group action.}

This is a Group Action. Let $x \in X$ and $\sigma \in H$. If $x = (g_1, g_2, \dots , g_p)$, observe that

\[ \sigma * x = (g_{\sigma(1)}, g_{\sigma(2)}, \dots, g_{\sigma(p)}). \]

Suppose $h(1) = n$. Then in general $h(i) = (i + n) \mbox{ mod }p.$ Therefore, we see that

\[ (g_{\sigma(1)}, g_{\sigma(2)}, \dots, g_{\sigma(p)}) = (g_{n}, g_{n+1}, \dots, g_p, g_1, \dots, g_{n-1}). \]

However, observe that

\[ g_1g_2\cdots g_p = g_1g_2 \cdots g_{n-1}g_n g_{n+1} \cdots g_p = e \implies (g_1g_2 \cdots g_{n-1})(g_n g_{n+1} \cdots g_p) = e. \]

Thus the elements $g_1g_2 \cdots g_{n-1}$ and $g_n g_{n+1} \cdots g_p$ in $G$ are inverses of each other. But know that if two group elements are inverses, either order of their product returns $e$. Therefore

\[ (g_ng_{n+1} \cdots g_{p})(g_1g_2 \cdots g_{n-1}) = g_ng_{n+1} \cdots g_{p}g_1g_2 \cdots g_{n-1} = e. \]

We therefore see that $(g_{n}, g_{n+1}, \dots, g_p, g_1, \dots, g_{n-1}) = \sigma *x \in X$.

Now we verify associativity. For any $\sigma_1, \sigma_2 \in H$, we see that

\[\begin{align*} \sigma_1 * \sigma_2*x &= \sigma_1 * (g_{\sigma_2(1)}, g_{\sigma_2(2)}, \dots, g_{\sigma_2(p)})\\ &= (g_{\sigma_1(\sigma_2(1))}, g_{\sigma_1(\sigma_2(2))}, \dots, g_{\sigma_1(\sigma_2(p))})\\ &= (\sigma_1 \sigma_2) * (g_1, g_2, \dots, g_p). \end{align*}\]

Thus $*$ is associative. Finally, if $\sigma$ is the trivial element,

\[ \sigma * x = (g_{\sigma(1)}, g_{\sigma(2)}, \cdots g_{\sigma(p)}) = (g_1, g_2, \dots, g_p) = x. \]

Therefore this is a group action.

\textcolor{NavyBlue}{Now that we've shown that this is a group action, we'll argue that the orbits are either of size 1 or $p$.}

The Orbits. For any $x \in X$ such that $x = (g_1, g_2, \dots, g_p)$, we see that the orbit $Hx$ will simply be all of the permutations of the $p$-tuple $(g_1, g_2, \dots, g_p)$. Note however that there are only $p$ many ways to rearrange this tuple, so that any orbit $Hx$ will be of size $p$.

Of course, the exception to this is if $g_1 = g_2 = \dots = g_p$. In this case, there are no other ways to reorganize the tuple. Hence the orbit will have size 1.

\textcolor{NavyBlue}{Finally, we will show that there exists a nontrivial orbit of size 1. This is equivalent to show that there exists a nontrivial element of $G$ of order $p$, which we'll eloaborate later.}

Orbit of Size 1. First let's count the elements of $X$. Observe that for any $(g_1, g_2, \dots, g_p) \in X$, the last element $g_p$ is always determined by the first $p-1$ elements. This is because if we know the first $p-1$ elements, then

\[ g_p = (g_1g_2 \cdots g_{p-1})^{-1} \]

in order for $g_1g_2\cdots g_p = e$. Since there are $|G|^{p-1}$ many ways to pick the first $p-1$ elements in any $p$-tuple of $X$, we see that $X = |G|^{p-1}$.

Now by hypothesis, $p$ divides $|G|$. Therefore $p$ divides $|X|$ so we may write $|X| = np$ for some integer $n$.

Since the orbits of $X$ form a partition, the orbits partition a set $np$ elements into orbits of size $1$ or size $p$. We know one orbit of size 1 exists (namely, the trivial orbit $He = \{(e, e, \dots, e)\}$), so there must exist at least $p-1$ nontrivial other orbits of size 1.

Let $Hx'$ be one of those orbits. Then for some $g \in G$ we have that $Hx = \{(g, g, \dots, g)\}$. However since $Hx \subset G$, what we have prove is the existence of a nontrivial element $g \in G$ such that $gg\cdots g = g^p = e$, which set out to show.

This completes the proof. Cacuhy's Theorem is an incredibly useful tool one can use in finite group theory. Here's an amazing and useful theorem who's proof is eased via Cauchy's Theorem.

Let $G$ be a group $p \ge 2$ a prime. If $|G| = p^n$ for some $n \in \mathbb{N}$, then $G$ has a subgroup of order $p^k$ for all $0 < k < n$. Note we didn't write $ 0 \le k \le n$. We could have, but we already know that there exists a subgroup of order $p^n$ (namely, $G$ itself) and that there exists a subgrou of order $p^0 = 1$ (namely, the trivial group).

\textcolor{NavyBlue}{To prove this, we'll use strong induction on the statement. Specifically, we'll induct on the powers of $n$.}

Let us induct on $n$ in the statement above. Then for $n = 1$, there is no such $k < n$. Hence the statement is vacuously true.

Next suppose that the statement is true up to order $p^n$, and let $G$ be a group of order $p^{n+1}$.

By Theorem 1.\ref{center_lemma}, we already note that $|Z(G)| > 1$ and hence is a multiple of $p$. By Cauchy's Theorem, we then know that $Z(G)$ contains an element $g$ of order $p$. Note that (1) $\left< g\right>$ is a subgroup of $Z(G)$ and (2) $h\left< g \right> = \left< g \right>h$ for all $h \in G$ (since, by definition of the center, every element of $Z(G)$ commutes with elements of $G$). Therefore $\left< g \right> \normal G$.

Let $H = \left< g \right>$. Since we just showed $H \normal G$ we can appropriately discuss the quotient group $G/H$.

Observe that $|G/H| = |G|/|H| = p^{n+1}/p = p^n$. \textcolor{purple}{Thus by hypothesis, $G/H$ has a subgroup of order $p^k$ for all $0 < k < n$. Denote these such subgroups of $G/H$ as}

\[ \{N_1/H, N_2/H, \cdots N_{n-1}/H\} \]

\textcolor{purple}{where $|N_k/H| = p^k$.} Since $H \normal G$, we know by the Fourth Isomorphism Theorem that every subgroup of $G/H$ is of the form $N/H$ where $H \le N \le G$. Thus we see that

\[ H \le N_k \le G \]

for all $0 < k < n$. But since $|N_k/H| = p^k$, and $|H| = p$, we see that each such $N_k$ will now have order $p^{k+1}$. Thus what we have shown is that $G$ itself contains subgroups of order $k$ for all $1 < k < n+1$. The subgroup $H$ of order $p$ is the final piece to this puzzle, and allows us to confirm that $G$ has a subgroup of order $p^k$ for all $0 < k < n$. By strong induction this holds for all $\mathbb{N}$, which completes the proof.