## § 7.2 Conjugacy: solved problems

15. Prove that if the center of the group G has index n, then every conjugacy class of G has at most n elements.

Solution: The conjugacy class of an element a in G has [G:C(a)] elements. Since the center Z(G) is contained in C(a), we have [G:C(a)] [G:Z(G)] = n. (In fact, [G:C(a)] must be a divisor of n.)

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16. Find all finite groups that have exactly two conjugacy classes.

Solution: Suppose that |G| = n. The identity element forms one conjugacy class, so the second conjugacy class must have n-1 elements. But the number of elements in any conjugacy class is a divisor of |G|, so the only way that n-1 is a divisor of n is if n = 2.

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17. Let G = D12, given by generators a,b with |a|=6, |b|=2, and ba=a-1b. Let H = { 1, a3, b, a3b }. Find the normalizer of H in G and find the subgroups of G that are conjugate to H.

Solution: The normalizer of H is a subgroup containing H, so since H has index 3, either NG (H) = H or NG (H) = G. Choose any element not in H to do the first conjugation.

aHa-1 = { 1, a(a3)a5, aba5, a(a3b)a5 } = { 1, a3, a2b, a5b }

This computation shows that a is not in the normalizer, so NG (H) = H. Conjugating by any element in the same left coset aH = { a, a4, ab, a4b } will give the same subgroup. Therefore it makes sense to choose a2 to do the next computation.

a2Ha-2 = { 1, a3, a2ba4, a2(a3b)a4 } = { 1, a3, a4b, ab }

Comment: It is interesting to note that an earlier problem shows that b, a2b, and a4b form one conjugacy class, while ab, a3b, and a5 b form a second conjugacy class. In the above computations, notice how the orbits of individual elements combine to give the orbit of a subgroup.

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18. Write out the class equation for the dihedral group Dn. Note that you will need two cases: when n is even, and when n is odd.

Solution: When n is odd the center is trivial and elements of the form ai b are all conjugate. Elements of the form ai are conjugate in pairs; am a-m since a2m 1. We can write the class equation in the following form.

|G| = 1 + ((n-1)/2) · 2 + n

When n is even, the center has two elements. (The element an/2 is conjugate to itself since it is equal to a-n/2. This shows that Z(G) = { 1, an/2 }.) Therefore elements of the form ai b split into two conjugacy classes. In this case the class equation has the following form.

|G| = 2 + ((n-2)/2) · 2 + 2 · (n/2)

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19. Show that for all n 4, the centralizer of the element (1,2)(3,4) in Sn has order 8 · (n-4)!. Determine the elements in the centralizer of ((1,2)(3,4)).

Solution: The conjugates of a = (1,2)(3,4) in Sn are the permutations of the form (a,b)(c,d). The number of ways to construct such a permutation is

n(n-1)/2 · (n-2)(n-3)/2 · 1/2 ,

and dividing this into n! gives the order 8 · (n-4)! of the centralizer.

We first compute the centralizer of a in S4. The elements (1,2) and (3,4) clearly commute with (1,2)(3,4). Note that a is the square of b = (1,3,2,4); it follows that the centralizer contains < b >, so b3 = (1,4,2,3) also belongs. Computing products of these elements shows that we must include (1,3)(2,4) and (1,4)(2,3), and this gives the required total of 8 elements.

To find the centralizer of a in Sn, any of the elements listed above can be multiplied by any permutation disjoint from (1,2)(3,4). This produces the required total |C(a)| = 8 · (n-4)!.

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