# Proving the image of a group homomorphism is a subgroup of its codomain

Can anyone help me prove this first step: given that $$\varphi : G \to H$$ is a group homomorphism, I seek to prove that $$\varphi(G)$$ is a subgroup of $$H$$.

I'm working on the First Isomorphism theorem and was wondering if someone could help me start off?

• What are φ, G, H? Apr 6, 2019 at 8:51
• What are $\varphi, G$ and $H$? Apr 6, 2019 at 8:51
• φ: G ----> H is a group homomorphism @Bernard Apr 6, 2019 at 8:52
• What do you need to show for something to be a subgroup? Apr 6, 2019 at 8:54
• Jonas referenced subgroup, i.e. the ordinary kind, not the "normal" kind, which are more special. Apr 6, 2019 at 9:05

Let $$\varphi : G \to H$$ be a group homomorphism. Then we want to show the image of $$\varphi$$ is a subgroup of $$H$$. Throughout, we will note that $$\ast$$ is assumed to be the operation of $$H$$ and $$\circ$$ that for $$G$$, just to avoid confusion.

Footnote: For some reason you stated that $$H$$ is a subgroup of $$G$$ in the comments of your question. I'm not sure why since this is not a necessary condition to my understanding.

Hopefully that $$\varphi(G),$$ if $$\varphi$$ is well-defined, would form a nonempty subset of $$H$$ is obvious enough. To show something is a subgroup, we need the properties of closure, identity, and inverses:

• Closure: Take any two elements in $$\varphi(G)$$ and show they multiply and give an element in $$\varphi(G)$$.
• Identity: Ensure $$\varphi(G)$$ has an identity element, i.e. $$\varphi(e)$$ where $$e$$ is the identity of $$G$$. Verify that it is indeed the identity.
• Inverses: Ensure each element of $$\varphi(G)$$ has an inverse in $$\varphi(G)$$.

Thus you need to show:

• Closure: $$\varphi(a),\varphi(b)\in\varphi(G) \implies\varphi(a) \ast \varphi(b) \in \varphi(G)$$ for all $$a,b \in G$$
• Identity: $$\varphi(e) \ast \varphi(a) = \varphi(a) \ast \varphi(e) = \varphi(a)$$ for all $$a \in G$$
• Inverses: $$\varphi(a)^{-1} \ast \varphi(a) = \varphi(a) \ast \varphi(a)^{-1} = \varphi(e)$$ for all $$a \in G$$, with each $$\varphi(a)^{-1}$$ existing for each $$\varphi(a)$$

Seems a little abstract but the proof largely makes use of the properties/definition of group homomorphisms, e.g. $$\varphi(a \circ b) = \varphi(a) \ast \varphi(b)$$.

• Thank you for this clear explanation. Apr 6, 2019 at 9:14
• This answer should go in the math.se hall of fame. Jan 10, 2021 at 19:48
• And as to the footnote: H doesn't need to be a subgroup of G since H need not even be a subset of G. Jan 10, 2021 at 19:50

I will use the one-step subgroup test.

Since $$e_H=\varphi(e_G)\in K:=\varphi(G)$$, $$K\neq\varnothing$$.

Since

$$K=\varphi (G)=\{h\in H\mid \exists g\in G, h=\varphi (g)\},$$

we have $$K\subseteq H$$.

Let $$x,y\in K$$. Then there exist $$a,b\in G$$ with $$x=\varphi(a),y=\varphi(b)$$. Thus

\begin{align} xy^{-1}&=\varphi(a)(\varphi(b))^{-1}\\ &=\varphi(a)\varphi(b^{-1})\tag{1}\\ &=\varphi(ab^{-1})\\ &\in K, \end{align}

where $$(1)$$ is justified here, and $$ab^{-1}\in G$$ as $$G$$ is a group.

Hence $$K\le H$$.