I'm confused that is a function which is a bijective homomorphism enough to prove the two groups are isomorphic, or we need all possible maps to be bijective homomorphism?
I saw a statement that even though a function $\varphi$ is not an isomorphism, it cannot ensure that the two groups are not isomorphic because there may exist other isomorphisms which hold. Is that true?
I try my best to make my question clear but I've already got myself messy so please forgive and help. Thanks!
Best Answer
Yes, if $\varphi: G\to H$ is a bijective group homomorphism from a group $G$ to a group $H$, then $G\cong H$; its inverse $\varphi^{-1}:H\to G$ is also a bijective homomorphism.
It is not enough to conclude $A\not\cong B$ for groups $A,B$, from just $\psi:A\to B$ not being a bijective homomorphism. For example, for $S_3$ (the symmetric group of degree three) and $D_3$ (the dihedral group of order six), we have $S_3\cong D_3$ despite
$$\begin{align} \theta:S_3 &\to D_3,\\ x&\mapsto e \end{align}$$
not being a bijective homomorphism (although it is a homomorphism).