Fun Exercise: Let $\varphi:G\to H$ be a surjective homomorphism of groups. If $G$ is abelian, then show that $H$ is abelian.

Question

kainui · Answer

I dunno what constitutes as proof. I guess the homomorphism takes $\varphi(g_1 * g_2) = \varphi(g_1) \otimes \varphi(g_2)$  where $*$ is the multiplication on G and $\otimes$ is the multiplication on H.
So I guess a homomorphism is saying $\varphi(g_1)=h_1$

Since multiplication is commutative in G,

$$\varphi(g_1*g_2)=\varphi(g_2*g_1)$$
then
$$\varphi(g_1)\otimes \varphi(g_2) = \varphi(g_2)\otimes \varphi(g_1)$$$$h_1\otimes h_2 = h_2 \otimes h_1$$
So the commutativity trickles through the homomorphism (assuming it is what I thought it was?)

across · Answer

That is correct. :-) And it follows that $H$ is abelian since $\varphi$ is surjective and this works for every $g\in G$. Here is your medal.

across · Answer

... every pair of... *

kainui · Answer

haha awesome. Ok so wait, is there some kinda nuance between Abelian and commutative? I've just been treating them as synonyms to be honest

across · Answer

I would say that they are equivalent, although the adjective "abelian" is reserved to groups whereas "commutative" is used more loosely.

kainui · Answer

Ahh ok I see I think I had only showed it for two specific members but maybe I should have called them $g_i$ and $g_j$ instead of $g_1$ and $g_2$ is that what you're referring to?

across · Answer

You were fine: You put no restrictions on $g_1$ and $g_2$, which implies that they are arbitrary. I was just being (arguably unnecessarily) pedantic with that statement. :-P

kainui · Answer

I've never taken abstract algebra so I sorta have vague notions of what happens. I don't know what the deal is with all these kinds of morphisms so I feel like I'm making it up and a bit disturbed that I had somehow given the right answer so easily. BUT maybe it's just supposed to be intuitive and I'm overthinking it. At any rate, I guess it's good to know I can sorta transfer this property between groups so thanks. :D

across · Answer

You did great. It looked to me like you had some experience with abstract algebra. :-)