The Unapologetic Mathematician

Mathematics for the interested outsider

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

7 Comments »

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