# Constructing an isomorphism to show two groups are isomorphic

For each $$n\geq 2$$, consider $$C_n = \{ (a,b) \in \mathbb{Z}^2 : a \equiv b \mod \: n\}$$. I want to show that $$C_n$$ is isomorphic to $$\mathbb{Z} \times \mathbb{Z}$$. To do this, I know I need to construct a bijection that preserves products as in the definition of an isomorphism. I’m at a loss for how to construct such a bijection, because I don’t see how any function I can think of can be both injective and surjective. Any guidance?

• Alternatively, you could avoid constructing an explicit isomorphism by using the result that any subgroup of ${\mathbb Z}^n$ is isomorphic to ${\mathbb Z}^m$ for some $m \le n$, and then rule out the cases $m=0,1$. – Derek Holt Feb 20 at 8:05

The bijection $$f:C_n\to\mathbb Z^2$$ may be constructed as follows: $$f(a,b)=\left(a,\frac{b-a}n\right)$$ Its inverse is $$f^{-1}(a,b)=(a,bn+a)$$
$$f$$ is a homomorphism because $$f(a,b)+f(c,d)=\left(a,\frac{b-a}n\right)+\left(c,\frac{d-c}n\right)=\left(a+c,\frac{(b+d)-(a+c)}n\right)=f(a+c,b+d)$$ $$f$$ is an injection: suppose $$f(a,b)=f(a',b')=(c,d)$$. Then $$a=a'$$, and manipulating $$\frac{b-a}n=\frac{b'-a'}n$$ we get $$b=b'$$ too. $$f$$ is surjective because of the inverse function demonstrated above. Thus $$f$$ is a bijection.
• I’m not following how $f\big((a_1,b_1),(a_2,b_2)\big) = f(a_1,b_1)f(a_2,b_2)$ using that function. – chris102212 Feb 20 at 1:26
• @chris102212 The question asks to show that $f$ is a bijection, not a homomorphism. Showing a homomorphism is trivial. – Parcly Taxel Feb 20 at 1:37
• @chris102212 Keep in mind that the group operation in both $\mathbb{Z}^2$ and $C_n$ is coordinatewise addition. – Daniel Schepler Feb 20 at 1:44
Look at the problem from a slightly different perspective: $$(a,b)\in C_n \iff n\mid(a-b)\iff b=a+nk$$ for some $$k\ge0$$. This means that the pair $$(a,k)$$ (where $$k=\frac{a-b}{n}$$) defines a bijection from $$\mathbb{Z}^2$$ to $$C_n$$ and it is not difficult to prove that it is also an isomorphism: indeed the map $$f(a,k)=(a,a+kn)$$ is a homomorphism since $$f((a,k)+(c,h))=f(a+c,k+h)=(a+c,a+kn+c+hn)=(a+c,a+c+kn+hn)=(a,a+kn)+(c,c+hn)=f(a,k)+f(c,h)$$.