This is Exercise 10 from Section 7: Groups and Homomorphisms, Chapter 1: Foundation, textbook Analysis I by Herbert Amann and Joachim Escher.

Let $(G,\odot)$ and $(H,\circledast)$ be groups, and let $$\varphi: G \times H \to G \space, \quad \langle g,h \rangle \mapsto g$$ be the projection onto the first coordinate.

  1. Show that $\varphi$ is a surjective homomorphism.

  2. Let $H' := \ker(\varphi)$. Show that $(G \times H)/H'$ and $G$ are isomorphic.

While the proof of 1. is quite easy, that of 2. is complicated to me. Please help me check it out. Thank you for your help!

My attempt:

In order to prove that $\varphi$ is a surjective homomorphism, $G \times H$ must be a group. It is confusing that the authors do not say anything about the operation on $G \times H$.

As a result, I guess $$\oplus: (G \times H) \times (G \times H) \to G \times H \space, \quad (\langle g_1, h_1 \rangle, \langle g_2, h_2 \rangle) \mapsto \langle g_1 \odot g_2, h_1 \circledast h_2 \rangle$$ is the operation on $G \times H$.

  1. $\varphi$ is a surjective homomorphism

First, $\varphi (\langle g_1, h_1 \rangle \oplus \langle g_2,h_2 \rangle) = \varphi (\langle g_1 \odot g_2, h_1 \circledast h_2 \rangle) = g_1 \odot g_2$. Second, $\varphi (\langle g_1, h_1 \rangle) \odot \varphi (\langle g_2,h_2 \rangle) = g_1 \odot g_2$. As a result, $$\varphi (\langle g_1, h_1 \rangle \oplus \langle g_2,h_2 \rangle) = \varphi (\langle g_1, h_1 \rangle) \odot \varphi (\langle g_2,h_2 \rangle)$$ and thus $\varphi$ is a homomorphism. Clearly, $\varphi$ is surjective.

  1. $(G \times H)/H'$ and $G$ are isomorphic

Since $\varphi$ is a homomorphism, $H'$ is a normal subgroup of $G \times H$. Then $(G \times H)/H'$ with the induced operation is a group, the quotient group of $G \times H$ modulo $N$.

$$\begin {array}{lrcl} & ((G \times H)/H') \times ((G \times H)/H') & \longrightarrow & (G \times H)/H' \\ & (\langle g_1, h_1 \rangle \oplus H' , \langle g_2, h_2 \rangle \oplus H') & \longmapsto & (\langle g_1, h_1 \rangle \oplus \langle g_2,h_2 \rangle) \oplus H' \end{array}$$

I will use the same symbol $\oplus$ for this induced operation.

It follows from $H' = \ker(\varphi)$ and $\varphi$ is a homomorphism that $$\langle g, h \rangle \oplus H' = \varphi^{-1}[\{\varphi(\langle g, h \rangle)\}] = \varphi^{-1}[\{g\}]$$ and that $$\langle g_1, h_1 \rangle \sim \langle g_2, h_2 \rangle \iff \varphi (\langle g_1, h_1 \rangle) = \varphi (\langle g_2, h_2 \rangle) \iff g_1 = g_2$$

Consider $$\begin {array}{lrcl} \psi : & (G \times H)/H' & \longrightarrow & G \\ & \langle g, h \rangle \oplus H' & \longmapsto & g \end{array}$$

Then $\psi$ is well-defined.

First, $\psi ((\langle g_1, h_1 \rangle \oplus H') \oplus (\langle g_2, h_2 \rangle \oplus H')) = \psi ((\langle g_1, h_1 \rangle \oplus \langle g_2,h_2 \rangle) \oplus H') = \psi (\langle g_1 \odot g_2, h_1 \circledast h_2 \rangle \oplus H') = g_1 \odot g_2.$

Second, $\psi (\langle g_1, h_1 \rangle \oplus H') \odot \psi (\langle g_2, h_2 \rangle \oplus H') = g_1 \odot g_2.$

As a result, $$\psi ((\langle g_1, h_1 \rangle \oplus H') \oplus (\langle g_2, h_2 \rangle \oplus H')) = \psi (\langle g_1, h_1 \rangle \oplus H') \odot \psi (\langle g_2, h_2 \rangle \oplus H')$$ and thus $\psi$ is a homomorphism. Clearly, $\psi$ is surjective.

If $\psi (\langle g_1, h_1 \rangle \oplus H') = \psi (\langle g_2, h_2 \rangle \oplus H')$, then $g_1 = g_2$. Hence $\varphi (\langle g_1, h_1 \rangle) = \varphi (\langle g_2, h_2 \rangle)$ and thus $\langle g_1, h_1 \rangle \sim \langle g_2, h_2 \rangle$. So $\langle g_1, h_1 \rangle \oplus H' = \langle g_2, h_2 \rangle \oplus H'$. As a result, $\psi$ is injective.

To sum up, $\psi$ is a bijective homomorphism and thus an isomorphism.

  • $\begingroup$ I posted an answer, but let me also recommend a source that teaches group theory how it should be taught: with the understanding of the objects that are transformed by the group. Use Part 1 of books.google.com/books/about/… $\endgroup$
    – avs
    Commented Mar 18, 2019 at 16:36
  • 3
    $\begingroup$ Your proof is correct, and pretty much mimics the proof of the first isomorphism theorem. In general, we have $G/\ker\phi\simeq\phi(G)$ for a homomorphism $\phi\colon G\to G'$ (with the isomorphism being $g(\ker\phi)\mapsto\phi(g)$) $\endgroup$
    – learner
    Commented Mar 18, 2019 at 16:39
  • $\begingroup$ Thank you so much! I think I need to learn to simplify notation in favor of brevity. $\endgroup$
    – Akira
    Commented Mar 18, 2019 at 16:48

2 Answers 2


The projection mapping $\phi:G\times H\rightarrow G:(g,h)\mapsto g$ is a surjective homomorphism. Note that the operation on $G\times H$ is point-wise: $$(g,h)\cdot (g',h') := (gg',hh').$$ The kernel of $\phi$ is $\{(e_G,h)\mid h\in H\}$, where $e_G$ is the unit element of $G$, and is isomorphic to $H$ by the projection mapping $\ker\phi \rightarrow H:(e_G,h)\mapsto h$.

Finally, the mapping $(G\times H)/\ker\phi \rightarrow G:\overline{(g,h)}\mapsto \phi(g,h)$ is an isomorphism, where $\overline{(g,h)}$ denotes the equivalence class of $(g,h)$ in the quotient group. This result is the socalled homomorphism theorem.


I think you can simplify the task by favoring prose over notation.

First, I will address the second part of the problem. Part 3 of the First isomomorphism theorem for groups (https://en.wikipedia.org/wiki/Isomorphism_theorems) says that, if $\psi : G_{1} \rightarrow G_{2}$ is a group homomorphism, then the image of $\psi$ is isomorphic to the factor space $$ G_{1} / \ker \psi. $$ Therefore, when the homomorphism is surjective (and I show below that this the case in your problem), its image is all of $G_{2}$, so $$ G_{2} \simeq G_{1} / \ker \psi. $$

To show that your homomorphism is surjective, just note that, for each $g \in G$, a pre-image under $\phi$ is found by taking any $h$ in $H$ and forming the ordered pair $(g, h)$. The projection of this pair onto the first coordinate is exactly $g$.


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