# Show that commutativity is a structural property

I am trying to prove that if two groups are isomorphic then one is abelian iff the other is abelian. This is a simple task, but I am a little confused about how to write it up.

Suppose that ##\phi : G \to H## is an isomorphism. Let ##a,b \in G##. Then ##ab = ba \implies \phi (a) \phi (b) = \phi (b) \phi (a)##. Here is where my question lies. To show that ##H## is abelian, I need to show that any two arbitrary elements commute. Why does ##\phi (a) \phi (b) = \phi (b) \phi (a)## tell me that any two arbitrary elements of ##H## commute? Does it have something to with ##\phi## being a bijection?

fresh_42
Mentor
I am trying to prove that if two groups are isomorphic then one is abelian iff the other is abelian. This is a simple task, but I am a little confused about how to write it up.

Suppose that ##\phi : G \to H## is an isomorphism. Let ##a,b \in G##. Then ##ab = ba \implies \phi (a) \phi (b) = \phi (b) \phi (a)##. Here is where my question lies. To show that ##H## is abelian, I need to show that any two arbitrary elements commute. Why does ##\phi (a) \phi (b) = \phi (b) \phi (a)## tell me that any two arbitrary elements of ##H## commute? Does it have something to with ##\phi## being a bijection?
Yes. ##\phi## covers all elements of ##H##:
##ab = ba \Longleftrightarrow \phi (a) \phi (b) = \phi (b) \phi (a) \Longleftrightarrow a'b'=b'a'##
You simply already have written all elements of ##H## since ##H=\operatorname{im} \phi = \{\,\phi(a)\,|\,a\in G\,\}##.

Mr Davis 97