Let $G$ be an abelian group, K a group and $f : G\rightarrow K $ a group homomorphism. Prove that $f (G)\subseteq K$ is an abelian subgroup of $K$.

abelian-groupsabstract-algebragroup-homomorphismgroup-theory

Let $G$ be an abelian group, K a group and $f : G\rightarrow K $ a group
homomorphism. Prove that $f (G)\subseteq K$ is an abelian subgroup of $K$.

I already proved that $f(G)$ is abelian, but I am not sure how to prove it is a subgroup of $K$. I need to show that it has the identity element, inverses, and closure. I am not sure how to show these 3 things

Best Answer

HINT:

  1. (Identity) If $e$ is the identity element for $G$, then $f(e)$ is the identity element for $f(G)$.

  2. (Inverse) $(f(a))^{-1} = f(a^{-1})$.

  3. (Closure) $f(a)f(b) = f(ab)$.

It's easier to use the subgroup test, but you will need the ideas above anyway.

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