# The Unapologetic Mathematician

## Conjugation

One of the most useful examples of a group acting on a set comes directly from group theory itself. Let $G$ be a group and $H$ be a subgroup of $G$. The subgroup $H$ acts on the set of all subgroups of $G$ as follows.

If $K$ is any subgroup of $G$ and $h$ is any element of $H$, then the set $hKh^{-1}$ of elements of $G$ of the form $hkh^{-1}$ with $k$ in $K$ is another subgroup of $G$. Indeed, if we take two elements $hkh^{-1}$ and $hk'h^{-1}$ of this set, their product is $hkh^{-1}hk'h^{-1}=hkk'h^{-1}$, which is again of the proper form since $kk'$ is in $K$. We call this subgroup the conjugation of $K$ by $h$.

Given two elements of $H$ we can check that $(hh')K(hh')^{-1}=hh'Kh'^{-1}h^{-1}$, so conjugating by $hh'$ is the same as conjugating by $h'$, then by $h$. That is, this defines an action of $H$ on the set of all subgroups of $G$.

Even better, $hKh^{-1}$ is not just another subgroup of $G$, it is isomorphic to $K$. In proving that $hKh^{-1}$ is a subgroup we showed that the function sending $k$ to $hkh^{-1}$ is a homomorphism. We can undo it by conjugating by $h^{-1}$, so it’s an isomorphism. We say that two subgroups of $G$ related by a conjugation are conjugate.

The subgroup of $H$ sending $K$ to itself — those $h$ in $H$ so that $hKh^{-1}=K$ — is called the normalizer of $K$ in $H$, written $N_H(K)$. We can verify that $K$ is a normal subgroup in $N_G(K)$, and that $K$ is normal in $G$ exactly when $N_G(K)=G$.

One orbit is particularly interesting to consider: $G$ is always sent to itself by conjugation. That is, given an element $h$ of $G$ the homomorphism sending $g$ to $hgh^{-1}$ is an automorphism of $G$. In fact, given any group $G$, the automorphisms of $G$ themselves form a group, called ${\rm Aut}(G)$. Conjugation gives us a homomorphism $c$ from $G$ to ${\rm Aut}(G)$: given an element $g$, $c(g)$ is the automorphism of conjugation by $g$.

We call automorphisms arising in this way “inner automorphisms”. The group ${\rm Inn}(G)$ of inner automorphisms on $G$ is the image of $c$ in ${\rm Aut}(G)$. If $g$ is an element of $G$ and $f$ is any automorphism of $G$, then $f\circ c(g)\circ f^{-1}$ is the automorphism sending $x$ in $G$ to

$f(gf^{-1}(x)g^{-1})=f(g)f(f^{-1}(x))f(g^{-1})=f(g)xf(g)^{-1}$

Which is just conjugation of $x$ by $f(g)$. This proves that ${\rm Inn}(G)$ is normal in ${\rm Aut}(G)$. The quotient ${\rm Aut}(G)/{\rm Inn}(G)$ is the group of outer automorphisms ${\rm Out}(G)$.

The kernel of $c$ is the set of elements $g$ so that $gg'g^{-1}=g'$ for all $g'$ in G. That is, for any $g'$ we have $gg'=g'g$, so the kernel of $c$ is the subgroup of $G$ consisting of elements that commute with every element of $G$. We call this subgroup the center of $G$.

Now, consider the group $S_n$ of permutations on $n$ letters. Determine how this group acts on itself by conjugation. Write out some conjugations in cycle notation to get an idea of what the answer should be.

February 22, 2007 - Posted by | Algebra, Group Actions, Group theory

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