Math 551, Assignment 1

Due Wednesday, September 18 in class

In these problems, G and H are groups. “Remarks” are for your information only and need not be proved.

1. Show that if ϕ : G → H is a homomorphism and g ∈ G has finite order n, then ϕ(g) has finite order dividing n. Moreover, the order of ϕ(g) equals n if and only if ⟨g⟩ ∩ ker(ϕ) = 1.

2. Suppose that X ≤ G and Y ≤ G. Define XY = {xy | x ∈ X, y ∈ Y }. Show that if X and Y are finite, then |XY | = |X||Y |/|X ∩ Y |. In addition, show (regardless of finiteness) that XY ≤ G if and only if XY = Y X.

3. If Z(G) = 1, then Z(Aut(G)) = 1.

4. Let λ : G → Σ_{G} be the mapping used in Cayley’s Theorem. Define ρ : G → Σ_{G} similarly: ρ(g)(x) = xg^{−1} for all g, x ∈ G. Show that C_{Σ_{G}}(ρ(G)) = λ(G). Here for any group X and subset Y ⊆ X, we define

C_{X}(Y) := {x ∈ X | xy = yx ∀y ∈ Y }.

5. Set d(G) := min{|S| | S ⊆ G and G = ⟨S⟩}, the “minimum number of generators of G”.

a) If H is a homomorphic image of G, then d(H) ≤ d(G). (Remark. It is far from true in general that if H ≤ G, then d(H) ≤ d(G), though it is true when d(G) = 1.)

b) If G and H are finite and gcd(|G|, |H|) = 1, then d(G × H) = max(d(G), d(H)).

6. Show that a group of exponent 2 is necessarily abelian, but that for any prime p > 2 there is a (triangular) subgroup of SL_{3}(Z/pZ) which has order p^{3} and exponent p.

7. Show that SL_{2}(Z) = ⟨ [ \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ] , [ \begin{matrix}0 & 1 \\ −1 & −1 \end{matrix} ] ⟩, and conclude that

P SL_{2}(Z) is generated by an element of order 2 and an element of order 3. (Note: P SL_{2}(Z) is the group of transformations of C ∪ {∞} of the form z ↦ (az + b)/(cz + d) with a, b, c, d ∈ Z and ad − bc = 1. The product of the two given matrices is a useful element.)

8. Let G = D_{8}, the symmetry group of the square. Let A = Aut(G). Show that |A| = 8. (Hint. A acts on G. Compute the orbit size and stabilizer size for some σ ∈ G of order 4.) Is A ≅ G?

9. Let α ∈ Aut(G), and define

C_{G}(α) = {g ∈ G | α(g) = g} and I_{G}(α) = {g ∈ G | α(g) = g^{−1}}.

Show that

a) C_{G}(α) ≤ G but I_{G}(α) need not be a subgroup of G

b) If |I_{G}(α)| > |G|/2, then α^{2} = 1

c) If I_{G}(α) = G, then G is abelian.

10. Continuing the previous problem, show that

d) If |I_{G}(α)| > 3|G|/4, then G is abelian and I_{G}(α) = G

e) If G is finite, α^{2} = 1 and |G| is odd, then |I_{G}(α)| = |G : C_{G}(α)|. (Hint. I_{G}(α) contains x^{−1}α(x) for every x ∈ G.)