Group Theory: Math30038, Sheet 1

GCS: Solutions

The course web site is available via http://www.bath.ac.uk/∼masgcs/

1. Let G be a group. Suppose that x ∈ G. Let

C_{G}(x) = {g ∈ G | gx = xg} ⊆ G.

Prove that C_{G}(x) ≤ G. Solution: 1 ∈ C_{G}(x) ≠ ∅. Moreover if g ∈ G and gx = xg, then xg^{−1} = g^{−1}x so if a, b ∈ C_{G}(x), then ab^{−1}x = axb^{−1} = xab^{−1} and therefore ab^{−1} ∈ C_{G}(x).

2. Let G be a group. Suppose that S ⊆ G. Let

C_{G}(S) = {g ∈ G | gs = sg∀s ∈ S}.

Prove that C_{G}(S) ≤ G. Solution: We have

C_{G}(S) = ∩_{s∈S}C_{G}(s)

and since the intersection of subgroups is a subgroup, we are done.

3. Let G be a group. Suppose that S ⊆ G. Let

N_{G}(S) = {g ∈ G | gS = Sg}

where gS = {gs | s ∈ S} and Sg = {sg | s ∈ S}. Prove that C_{G}(S) ≤ N_{G}(S) ≤ G. Solution: The fact that N_{G}(S) is a group follows the outline of the proof that C_{G}(x) is a group. The condition to be in C_{G}(S) is stronger than that to be in N_{G}(S) so C_{G}(S) ≤ N_{G}(S) ≤ G.

4. Let G be a group. Suppose that S ⊆ G and that x ∈ N_{G}(S). Prove that C_{G}(S)x = xC_{G}(S). Solution: We will show that C_{G}(S) = x^{−1}C_{G}(S)x which will suffice. Suppose that s ∈ S, then xs = s′x for some s′ ∈ S. Now shoose any c ∈ C_{G}(S), then

x^{−1}cxs = x^{−1}cs′x = x^{−1}s′cx = x^{−1}(xsx^{−1})cx = sx^{−1}cx

and therefore sx^{−1}c ∈ C_{G}(S) and so x^{−1}C_{G}(S)x ≤ C_{G}(S). Replacing x by x^{−1} in this argument yields that xC_{G}(S)x^{−1} ≤ C_{G}(S). Premultiplying by x^{−1} and postmultiplying by x gives C_{G}(S) ≤ x^{−1}C_{G}(S)x. Now we have two mutually reverse inclusions so C_{G}(S) = x^{−1}C_{G}(S)x for all x ∈ N_{G}(S).

5. Let G be a finite group. Suppose that ∅ ≠ H ⊆ G has the property that if a, b ∈ H, then ab ∈ H. Does it follow that H ≤ G? What happens if we relax the condition that G is finite? Solution: If h ∈ H, then ⟨h⟩ ≤ H and so o(h) is a natural number n. If h = 1 then h^{−1} = h ∈ H. Otherwise n > 1 and then h^{−1} = h^{n−1} ∈ H so H is a subgroup of G. In the event that G is infinite, things fall apart. For example perhaps G = ℤ under addition, the ℕ is a non-empty additively closed subset which is not a subgroup since −1 ∉ ℕ.

6. Suppose that G is a group and that x ∈ G. Prove that (x^{−1})^{−1} = x. Solution: x^{−1}x = 1 = x^{−1}(x^{−1})^{−1}. Premultiplying by x gives the result.

7. Does there exist a group G containing elements a, b such that a^{2} = b^{2} = (ab)^{3} = 1? Solution: Yes, the trivial group will do it. More interestingly, let G be the group S_{3}, and put a = (1, 2) and b = (2, 3). Now the orders of a, b and abare exactly “as advertized”.

8. Suppose that G is a group with the property that x^{2} = 1 whenever x is an element of G. Show that G must be abelian. Solution: Suppose that a, b ∈ G, then abab = 1 = baab. Postmultiplying by ba yields ab = ba.

9. (Challenge) Suppose that G is a group with the property that x^{3} = 1 whenever x is an element of G. Show that G need not be abelian. Solution: Consider the set of 3 by 3 upper triangular matrices with entries in F = ℤ_{3}, the field of integers modulo 3 which have 1s on the leading diagonal. It is easy to verify that these 27 matrices form a group. Each matrix is of the form I + Δ wheer I is the 3 by 3 identity matrix and Δ is strictly upper triangular. Note that Δ^{3} is the zero matrix, so the inverse of I + Δ is I − Δ + Δ^{2} and moreover (I + Δ)^{3} = I + 3Δ + 3Δ^{2} = I so ev ery element of the group has order dividing 3. The matrices I + E_{12} and I +E_{23} do not commute as may be verified by direct calculation; here E_{ab} is the 3 by 3 matrix with entries in F where the entrie in the i-th row and j-th column is δ_{ia}δ_{jb} where δ is the Kronecker delta.