# Let $G,H$ be groups and $\varphi:G \times H\to G$ and $H'=\ker(\varphi)$. Show that $(G\times H)/H'\cong G$

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.

• 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/…
– avs
Mar 18 '19 at 16:36
• 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 $g(\ker\phi)\mapsto\phi(g)$) Mar 18 '19 at 16:39
• Thank you so much! I think I need to learn to simplify notation in favor of brevity. Mar 18 '19 at 16:48

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.
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 $$g$$.