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?

  • $\begingroup$ What are φ, G, H? $\endgroup$
    – Bernard
    Apr 6, 2019 at 8:51
  • $\begingroup$ What are $\varphi, G$ and $H$? $\endgroup$
    – Jonas Lenz
    Apr 6, 2019 at 8:51
  • $\begingroup$ φ: G ----> H is a group homomorphism @Bernard $\endgroup$
    – Sammy.d
    Apr 6, 2019 at 8:52
  • 1
    $\begingroup$ What do you need to show for something to be a subgroup? $\endgroup$
    – Jonas Lenz
    Apr 6, 2019 at 8:54
  • 1
    $\begingroup$ Jonas referenced subgroup, i.e. the ordinary kind, not the "normal" kind, which are more special. $\endgroup$ Apr 6, 2019 at 9:05

2 Answers 2


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. $\varphi(a \circ b) = \varphi(a) \ast \varphi(b)$.

  • $\begingroup$ Thank you for this clear explanation. $\endgroup$
    – Sammy.d
    Apr 6, 2019 at 9:14
  • $\begingroup$ This answer should go in the math.se hall of fame. $\endgroup$
    – Hank Igoe
    Jan 10, 2021 at 19:48
  • 1
    $\begingroup$ 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. $\endgroup$
    – Hank Igoe
    Jan 10, 2021 at 19:50

I will use the one-step subgroup test.

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


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

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 $K\le H$.


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