Mathematics 243, section 1 – Algebraic Structures Selected Solutions – Problem Set 7 November 20, 2003

3.4/1. Let ϕ : G → H be a group isomorphism. Then we know ϕ is 1-1, onto and satisfies the group homomorphism property ϕ(xy) = ϕ(x)ϕ(y) for all x, y ∈ G. Since ϕ is 1-1 and onto, there exists an inverse mapping ϕ^{−1} : H → G. Since ϕ = (ϕ^{−1})^{−1}, ϕ^{−1} is also oneto-one and onto. Hence we only need to show that ϕ^{−1} is also a group homomorphism. Let w, z ∈ H. Since ϕ is onto, there exist x, y ∈ G such that ϕ(x) = w and ϕ(y) = z (so ϕ^{−1}(w) = x and ϕ^{−1}(z) = y. Hence

wz = ϕ(x)ϕ(y) = ϕ(xy) ⇒ ϕ^{−1}(wz) = xy = ϕ^{−1}(w)ϕ^{−1}(z)

This is what we wanted to show.

3.4/2. Although the notation ϕ_{2}ϕ_{1} is slightly ambiguous in the problem, the only consistent interpretation from the context is that this is the composite mapping:

ϕ_{2} ∘ ϕ : G_{1} → G_{3}

From work in Chapter 1, we know that the composition of 1-1 mappings is 1-1 and the composition of onto mappings is onto. Hence since ϕ_{1} and ϕ_{2} are group isomorphisms, ϕ_{2} ∘ ϕ_{1} is also 1-1 and onto. Then, if x, y ∈ G_{1} are arbitrary, we have

(ϕ_{2} ∘ ϕ_{1})(xy) = ϕ_{2}(ϕ_{1}(xy)) = ϕ_{2}(ϕ_{1}(x)ϕ_{1}(y)) = ϕ_{2}(ϕ_{1}(x))ϕ_{2}(ϕ_{1}(y))

since both ϕ_{1} and ϕ_{2} have the group homomorphism property. It follows that ϕ_{2} ∘ ϕ_{1} is also an isomorphism of groups.

3.4/21. Let G and H be any two cyclic groups of order n. By definition, this means that G = ⟨a⟩, H = ⟨b⟩ for some a ∈ G and b ∈ H, and moreover, a^{n} = b^{n} = e and n is the smallest strictly positive integer with this property. Therefore,

\begin{aligned}G & = \{e = a^{0}, a = a^{1}, …, a^{n−1}\} \\ H & = \{e = b^{0}, b = b^{1}, …, b^{n−1}\} \end{aligned}

Define ϕ : G → H by ϕ(a^{k}) = b^{k} for all k ∈ Z. Then ϕ is clearly one-to-one and onto since it maps distinct elements of G to distinct elements of H and every element of H is the image of some element of G. Finally, letting x = a^{ℓ} y = a^{m} be any two elements of G, we have

ϕ(a^{ℓ} · a^{m}) = ϕ(a^{ℓ+m}) = b^{ℓ+m} = b^{ℓ} · b^{m} = ϕ(a^{ℓ})ϕ(a^{m})

so ϕ is an isomorphism of groups.

3.5/10. Let H be a homomorphic image of G (that is, there exists an onto group homomorphism ϕ : G → H). Let w, z be arbitrary elements of H. Then since ϕ is onto, w = ϕ(x) and z = ϕ(y) for some x, y in G. We have, using commutativity in G:

wz = ϕ(x)ϕ(y) = ϕ(xy) = ϕ(yx) = ϕ(y)ϕ(x) = zw

Hence H is also abelian.