I.15.1

G is a topological group. G_{0} is the connected component of G, containing the identity element, e. Show that G_{0} is a closed, normal subgroup of G. Proof Connected components are always closed. Let x ∈ G_{0}. Since multiplication and group inversion are continuous, then G_{0}x^{−1} is connected. Since e = xx^{−1} ∈ G_{0}x^{−1}, then G_{0}x^{−1} ⊂ G_{0}. So G_{0} is a subgroup of G. We have that zG_{0}z^{−1} is connected, for all z ∈ G, since multiplication and group inversion are continuous. Now e = zez^{−1} ∈ zGz^{−1}, so zGz^{−1} ⊂ G, hence G_{0} ◁ G.

I.15.2

ϕ : G → H is an onto homomorphism. G, H are topological groups.

a) Show that K := Kerϕ is a closed, normal subgroup of G. Proof ϕ is continuous. Since {e} ⊂ H is closed in H, then K = Kerϕ = ϕ^{−1}(e) is closed. Let x, y ∈ K. We have

ϕ(xy^{−1}) = ϕ(x)ϕ(y^{−1}) = eϕ(y)^{−1} = ee^{−1} = e

so Kerϕ is a subgroup of G.

We want to prove that zKz^{−1} ⊂ K, for all z ∈ G Let z ∈ G, x ∈ Kerϕ. We then have:

ϕ(zxz^{−1}) = ϕ(z)ϕ(x)ϕ(z^{−1}) = ϕ(z)eϕ(z^{−1}) = ϕ(z)ϕ(z^{−1}) = ϕ(zz^{−1}) = ϕ(e) = e,

hence K ◁ G.

b) Assume that G is compact. Show that G/K ≈ H.

Proof

By proposition 15.12, G/K is a topological group and by proposition 15.13, the canonical mapping, π : G → G/K is open and continuous. Let’s define ψ : H → G/K with ψ(x) = πϕ^{−1}(x). The mapping is well defined by definition of π. It is also clearly bijective. If we take an open set U ⊂ H, then πϕ^{−1}(U) is open in G/K, because ϕ is continuous, and π is an open mapping, hence π^{−1} is continuous. So ψ is continuous. Since G is compact, ψ^{−1} is also continuous by Theorem 7.8. So we have proved that G/K ≈ H.

I.15.3

G is a compact topological group. g ∈ G. A := {g^{n}|n = 0, 1, 2, … }.

a) Show that \overline{A} is a subgroup of G.

Proof

Since \overline{A} is a closed subset of a compact subset, it is compact. Therefore, every sequence in \overline{A}, has a convergent subsequence. Let’s look at the sequence

{g^{n}}_{n=0}^{∞}

and let

{g^{n_{k}}}_{n=0}^{∞}

be a convergent subsequence. Let’s define

h_{i} = lim_{k→∞, k⩾i}g^{−n_{i}} · g^{n_{k}}

Now

g^{n_{i}} · h_{i} = g^{n_{i}} lim_{k→∞, k⩾i}g^{−n_{i}} · g^{n_{k}} = lim_{k→∞, k⩾i}g^{n_{k}} = h

We also have that

lim_{i→∞} h_{k}^{−1} = lim_{i→∞} h^{−1}g^{n_{i}} = h^{−1} lim_{i→∞} g^{n_{i}} = h^{−1}h = e

so lim_{i→∞} h_{i} = e. Therefore e is an accumulation point in A. That is e ∈ \overline{A}. By using the same method on the set g^{i}A, we have that e ∈ \overline{g^{i}A}, ∀i ∈ ℕ. Hence g^{−i} ∈ \overline{A}, ∀i ∈ ℕ, and hence \overline{A} is a subgroup of G.

b) Is this true if G is not compact? Solution (If you find a counterexample, stick to it, even though you find a simpler one later) Let G = Gl(n, ℝ), and let

g = [\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}]

Since

[\begin{matrix}1 & p \\ 0 & 1 \end{matrix}] · [\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}] = [\begin{matrix}1 & p + 1 \\ 0 & 1 \end{matrix}]

A is a discrete subset in G, so \overline{A} = A and g^{−1} ∉ G. Hence \overline{A} is not a subgroup of G. We have therefore showed, that compactness is necessary.

I.15.4

G is a compact topological group.

Show that every neighborhood of e contains a neighborhood V which is invariant under conjugation.

One idea, which doesn’t seem to work

Let U be a neighborhood of e. Define

V = {h ∈ U|ghg^{−1} ∈ U, ∀g ∈ G}

Now, e ∈ V because geg^{−1} = e ∈ U, ∀g ∈ G. Since group multiplication and group inversion are continuous, V is a neighborhood of e contained in U (??). We want to show that gV g^{−1} ⊂ V, ∀g ∈ G.

Let h ∈ V, g ∈ G. Since h ∈ V, ghg^{−1} ∈ O, ∀g ∈ G. ∀g_{1} ∈ G we have that

g_{1}(ghg^{−1})g_{1}^{−1} = (g_{1}g)h(g_{1}g)^{−1} ∈ O

because g_{1}g ∈ G and h ∈ V. So gV g^{−1} ⊂ V, ∀g ∈ G, hence gV g^{−1} = V, ∀g ∈ G, because G is a group.

I haven’t succeeded in showing that V is a neighborhood of e. Therefore this problem is still unsolved on my behalf.