1

I am completing this question as part of my study for my upcoming mid-term:

Find an isomorphism between (C, +) and (F, o). Justify your answer

  • C is the set of complex numbers
  • ∀a ∈ C, the map fa : C → C given by f(z) = z+a namely the translation by a
  • Collection of maps F is defined: enter image description here

The question I am trying to answer above is part c) of a three-part question. Parts a and b asked to show (C, +) and (F, o) were groups and I did this by showing they followed the axioms of closure, associativity, identity element, and inverse to which I concluded both were in fact groups.

I am unsure how to find an isomorphism between the two groups any help would be appreciated, thanks in advance!

Renee Ofadu
  • 63
  • 2
    Try mapping $a$ to $f_a.$ – Chris Leary Oct 29 '22 at 21:30
  • @ChrisLeary Thanks for the response I know I need to make some function g that takes a complex number input and its output is another function fa, where fa = f(z) = z + a, and that function g is supposed to be our isomorphic function (i.e. mapping a to fa) but I am not sure how to do this or where to begin? :) – Renee Ofadu Oct 29 '22 at 21:33
  • Going off of what Chris said above, why not let $g:\mathbb{C}\to\mathcal{F}$ be the function which maps $a$ to $f_a$. How can you check that this is the isomorphism you want? What properties does $g$ need to satisfy? – ondwats Oct 29 '22 at 21:45
  • @ondwats would it be correct to say I want to check that g is bijective and a homomorphism for me to know this is the isomorphism I want (i.e. g needs to be bijective and a homomorphism)? I understand this much but I am unsure how to show bijectivity and homomorphism. Thanks for your input also :) – Renee Ofadu Oct 29 '22 at 21:49
  • Yup; an isomorphism is a bijective homomorphism. So you'd just need to check infectivity, surjectivity, and that $g(a+b) = f_a∘ f_b$. – ondwats Oct 29 '22 at 21:57

1 Answers1

1

Consider $\phi: \mathbb{C} \to \mathcal{F}_\mathbb{C}$ defined trivially: $\phi(a) = f_a$. You should convince yourself that such a function is well-defined.

We must now show that $\phi$ an isomorphism.

First, we show that $\phi$ is a bijection. Given $\phi(a_1) = \phi(a_2)$, it means that the function $f_{a_1} = z + a_1$ is equal to the function $f_{a_2} = z + a_2$. Of course, this implies $a_1 = a_2$, so $\phi$ is injective. Next, take an arbitrary element of $\mathcal{F}_\mathbb{C}$, $f_b = z + b \implies b \in \mathbb{C}$. Note that $\phi(b)$ is precisely $f_b$, and since $f_b$ was arbitrary, $\phi$ must be surjective.

Next, we show that $\phi$ is a homomorphism. For this, we need to show that $\phi (a_1 + a_2) = \phi (a_1) \circ \phi (a_2)$ for arbitrary $a_1, a_2 \in \mathbb{C}$. Observe that $\phi (a_1 + a_2) = f_{a_1 + a_2} = z + a_1 + a_2$, and $\phi (a_1) \circ \phi (a_2) = f_{a_1} \circ f_{a_2} = (z + a_2) + a_1 = z + a_1 + a_2$. They are equal, so $\phi$ preserves the group structures and is a homomorphism.

K. Jiang
  • 7,210