I Show that commutativity is a structural property

1,456
44
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
Insights Author
2018 Award
10,074
6,806
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\,\}##.
 

Want to reply to this thread?

"Show that commutativity is a structural property" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top