# Injective homomorphisms and subgroups

I've just started self-learning group theory, and I came accros Cayley's theorem. Which made me wondering about the following question: Given two groups $G$ and $H$, and there exist an injective homomorphism $f: G \rightarrow H$, then $G$ is isomorphic to a subgroup of $H$.

My attempt: Let $K$ be the range of the homomorphism. Then $f: G \rightarrow K$ is surjective, but it is also injective, so bijective. So $G$ and $K$ are isomorphic. But how do I know that $K$ is a subgroup? I think it is enough that since $K$ is isomorphic to a group $G$, $K$ must be a group, but why a subgroup of $H$. Sure, the set of $K$ is a subset of the set of $G$, but this not a proof. Can someone give the correct argument?

• More generally, regardless of whether a homomorphism is injective, its image will always be a group. In fact, $\operatorname{Im}(f) \cong G/ \ker(f)$. Commented Jul 10, 2015 at 7:41
• @KajHansen What does the / mean here? I know what the image and the kernel are. Commented Jul 10, 2015 at 7:44
• The "$/$" means "quotient group" (or "factor group"). Commented Jul 10, 2015 at 7:48
• @KajHansen what you said is true iff it is surjective homomorphism.. isnt it? it cant be always. Commented Aug 14, 2015 at 6:50

For the neutral element: $e_H = f(e_G) \in K$.
For the inverse: for $y \in K$, let $x \in G$ with $f(x) = y$. Then $y^{-1} = f(x)^{-1} = f(x^{-1}) \in K$.
For the product: let $y, y' \in K$. Let $x, x' \in G$ with $f(x) = y, \space f(x') = y'$. Then $y y' = f(x) f(x') = f(x x') \in K$.
Let $H,K$ be groups. We call $K$ a subgroup of $H$ if it is a subset and if its group operation is the group operation of $H$ restricted to $K$. You showed that $K$ is a group (isomorphic to $G$) , it is clearly a subset of $H$, and of course the group operation of $K$ is the restriction of that of $H$ (that is what "$f$ is a homomorphism" implies).
More generally, for any group homomorphism $f\colon G\to H$, we have that the image $K=f(G)$ is a subgroup. Instead of using the definition we use the well-known criterion: a subset $K$ of $H$ is a subgroup if it is not empty and $a,b\in K$ implies $ab^{-1}\in K$. Being the image of the nonempty set $G$, clearly $K$ is not empty. Given $a,b\in K$ we find $x,y\in G$ with $f(x)=a$, $f(y)=b$. Then $ab^{-1}=f(x)f(y)^{-1}=f(xy^{-1})\in K$.