Surjective group homomorphisms preserve abelianicity

If \varphi : G \rightarrow H is an isomorphism, prove that G is abelian if and only if H is abelian. If \varphi : G \rightarrow H is only a homomorphism, what additional conditions on \varphi (if any) are sufficient to ensure that if G is abelian, then so is H?

Let \varphi : G \rightarrow H be a group isomorphism.

(\Rightarrow) Suppose G is abelian, and let h_1,h_2 \in H. Since \varphi is surjective, there exist g_1, g_2 \in G such that \varphi(g_1) = h_1 and \varphi(g_2) = h_2. Now we have h_1 h_2 = \varphi(g_1) \varphi(g_2) = \varphi(g_1 g_2) = \varphi(g_2 g_1) = \varphi(g_2) \varphi(g_1) = h_2 h_1. Thus h_1 and h_2 commute; since h_1, h_2 \in H were arbitrary, H is abelian.

(\Leftarrow) Suppose H is abelian, and let g_1, g_2 \in G. Then we have \varphi(g_1 g_2) = \varphi(g_1) \varphi(g_2) = \varphi(g_2) \varphi(g_1) = \varphi(g_2 g_1). Since \varphi is injective, we have g_1 g_2 = g_2 g_1. Since g_1, g_2 \in G were arbitrary, G is abelian.

Note that in the proof of (\Rightarrow) above, we only needed to use the surjectivity of \varphi. So in fact the image of any surjective group homomorphism from an abelian group is abelian. Similarly, in the proof of (\Leftarrow) above we only used injectivity; thus the source of any injective group homomorphism to an abelian group is abelian.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: