Prove $C_2\times C_3$ is isomorphic to $C_6$

Question

I am struggling to understand whether or not this proof is legitimate, here is my attempt:

Prove that C2 × C3 is isomorphic to C6

Define $\varphi:C_6\rightarrow C_2\times C_3$ by the mapping $\varphi(c^k)=(a^k,b^k)$, where $c,a,b$ generate $C_6,C_2,C_3$ respectively. Then;

$\begin{align} \varphi(c^k\cdot c^l)&=\varphi(c^{k+l})\\&=(a^{k+l},b^{k+l})\\&=(a^k,b^k)\cdot(a^l,b^l)\\&=\varphi(c^k)\varphi(c^l)\end{align}$

Define $\psi:C_2\times C_3\rightarrow C_6$ by the map $\psi(a^k,b^k)=c^k$. Clearly $\varphi(\psi(a^k,b^k))=(a^k,b^k)$ and $\psi(\varphi(c^k))=c^k$, so $C_2\times C_3 \cong C_6$. Or perhaps it is clearer to explicitly construct the bijection:
$\begin{array}\\ &e\mapsto(e,e)\\&c\mapsto(a,b)\\&c^2\mapsto(e,b^2)\\&c^3\mapsto(a,e)\\&c^4\mapsto(e,b)\\&c^5\mapsto(a,b^2) \end{array}$

My apologies if this is blatantly wrong, any advice is greatly appreciated.

Answer 1

Define $\psi\colon C_2\times C_3\to C_6$ by the map $\psi(a^k,b^k)=c^k$.

This is incomplete. It is not immediately clear that all elements in $C_2\times C_3$ look like $(a^k,b^k)$. What about $(a^1,b^2)$?

The explicit-writing-out-stuff way is correct. It proves completely that it is a bijection. (And then of course you notice that all elements of $C_2\times C_3$ actually look like $(a^k,b^k)$.)

Hope this helps. :)