MATH 351 – Modern Algebra I Selected Solutions for Problem Set 10 December 7, 2018

7.18. Let G be an infinite group and let H be the subset of G consisting of the elements with only finitely many conjugates. Equivalently H is the subset of a ∈ G such that |C_{a}| is finite. We have to show that H is a subgroup of G.

• First, we consider e. For all g ∈ G, geg^{−1} = e. Therefore C_{e} = {e} and |C_{e}| = 1. Since this is finite, e ∈ H.

• Now let a, b ∈ H and consider C_{ab}. For all g ∈ G, we have using associativity

gabg^{−1} = ga(g^{−1}g)bg = (gag^{−1})(gbg^{−1}).

The factor on the left is an element of C_{a} and the factor on the right is an element of C_{b}. Hence when we form these products, we obtain at most |C_{a}| · |C_{b}| different elements of G. There could be also strictly fewer than |C_{a}| · |C_{b}| in some cases, but in any case, the number |C_{ab}| is finite. This shows ab ∈ H.

• Now let a ∈ H and consider |C_{a^{−1}}|. The elements of this class all have the form ga^{−1}g^{−1} for g ∈ G. Note that the inverse of this element of G is

(ga^{−1}g^{−1})^{−1} = gag^{−1},

which is one of the elements of |C_{a}|. Since we began by assuming a ∈ H, there are only finitely many different values obtained for gag^{−1}. That implies there are only finitely many conjugates of a^{−1} too since each element of G has a unique inverse in G. That shows a^{−1} ∈ H.

Putting together these three points, we get that H is a subgroup of G.

7.24. Most people saw correctly that you want to use Sylow II to say that any two Sylow 2-subgroups H_{1} and H_{2} must be conjugate. But then we must say why (or prove directly that) if H_{2} = gH_{1}g^{−1} for two subgroups H_{1}, H_{2} of G and some element g ∈ G, then H_{1} and H_{2} must be isomorphic as groups. Here’s how to show that. Consider the conjugation mapping α:

\begin{matrix}α : H_{1} & → H_{2} \\ h & ↦ ghg^{−1} \end{matrix}

We claim that α is an isomorphism. First, it is a group homomorphism because if h, k ∈ H_{1}, then

α(hk) = ghkg^{−1} = gh(g^{−1}g)hg^{−1} = (ghg^{−1})(gkg^{−1}) = α(h)α(k).

The mapping α is surjective with image H_{2} by Sylow II. Moreover, since |H_{1}| = p^{n} = |H_{2}| if |G| = p^{n}r with gcd(p, r) = 1, once we know α is surjective, then α must be injective as well. Hence α is an isomorphism of groups.

But this shows that it is impossible for one Sylow 2-subgroup H_{1} ≅ Z_{4} and another Sylow 2-subgroup H_{2} ≅ Z_{2} × Z_{2}. The groups Z_{4} and Z_{2} × Z_{2} are not isomorphic, by the fundamental theorem of finite abelian groups.

7.30. Let |G| = 56 = 2^{3} · 7. We claim that G must have a normal subgroup other than {e} and G, so G is not simple. Let p = 7. By Sylow III, the number of Sylow 7-subgroups is ≡ 1 mod 7 and that number divides 8. Thus both 1 and 8 Sylow 7-subgroups are possible. If there is just one, then as usual, Sylow II implies that subgroup is normal and we are done. On the other hand, if there are 8 Sylow 7-subgroups in G, each of order = 7, then since 7 is prime, each of the Sylow 7-subgroups is cyclic and there are 6 elements of order 7 in each of them. Distinct Sylow 7-subgroups can only intersect in the identity element. Counting up the total number of elements in the union of those subgroups we get 1 + 8 · 6 = 49. This leaves only 7 other elements of G, and those 7 elements together with e make up exactly one subgroup of order 8 by Sylow I (there exists at least one subgroup with 2^{3} = 8 elements, but there is no room for more than one in this case(!)). Hence if there are 8 Sylow 7-subgroups, there must be only one Sylow 2-subgroup, hence that Sylow 2-subgroup is normal by Sylow II. Hence G is not simple in this case either.