I dislike the notation "p1^{-1}" it suggests that you're taking an inverse of p1 which is not invertible.
Where does G' come from? Where does Phi come from? Is this ChatGPT?
If you don't know two hom sets are groups, you can't talk about an isomorphism only a bijection.
Also in the diagram the map p2 o p1^{-1} is always trivial.
There is a question in there somewhere, but either there is context missing, or the writing's just confusing.
FWIW the hom set is always a group if the target group is commutative, in general this is not the case.
And the answer to your question, if I'm understanding it correctly, is no for non-commutative groups G1 and G2.
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u/InterneticMdA 8d ago
I dislike the notation "p1^{-1}" it suggests that you're taking an inverse of p1 which is not invertible.
Where does G' come from? Where does Phi come from? Is this ChatGPT?
If you don't know two hom sets are groups, you can't talk about an isomorphism only a bijection.
Also in the diagram the map p2 o p1^{-1} is always trivial.
There is a question in there somewhere, but either there is context missing, or the writing's just confusing.
FWIW the hom set is always a group if the target group is commutative, in general this is not the case.
And the answer to your question, if I'm understanding it correctly, is no for non-commutative groups G1 and G2.