**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 = D_{12},
given by generators a,b with |a|=6, |b|=2, and ba=a^{-1}b.
Let H = { 1, a^{3}, b, a^{3}b }.
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 N_{G} (H) = H or N_{G} (H) = G.
Choose any element not in H to do the first conjugation.

aHa^{-1}
= { 1, a(a^{3})a^{5},
aba^{5}, a(a^{3}b)a^{5} }
= { 1, a^{3}, a^{2}b, a^{5}b }

This computation shows that a is not in the normalizer,
so N_{G} (H) = H.
Conjugating by any element in the same
left coset aH = { a, a^{4}, ab, a^{4}b }
will give the same subgroup.
Therefore it makes sense to choose a^{2} to do the next computation.

a^{2}Ha^{-2}
= { 1, a^{3}, a^{2}ba^{4},
a^{2}(a^{3}b)a^{4} }
= { 1, a^{3}, a^{4}b, ab }

*Comment:*
It is interesting to note that an earlier problem shows
that b, a^{2}b, and a^{4}b form one conjugacy class,
while ab, a^{3}b, and a^{5} 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 D_{n}.
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 a^{i} b are all conjugate.
Elements of the form a^{i} are conjugate in pairs;
a^{m} a^{-m}
since a^{2m} 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 a^{n/2} is conjugate to itself since
it is equal to a^{-n/2}.
This shows that Z(G) = { 1, a^{n/2} }.)
Therefore elements of the form a^{i} 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 S_{n}
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 S_{n} 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 S_{4}.
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 b^{3} = (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 S_{n},
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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