I've read in three different books about groups, subgroups and homomorphism but I just can't get the concept. I don't find the examples clear enough. For example, why isn't $$\Phi: \mathbb{Z} \rightarrow \mathbb{Q}, x \rightarrow x^2$$ a homomorphism?

I have to the do the following exercise:

Let $(G, \bullet)$ and $(H, *)$ be groups, where the neutral elements are $e_G$ and $e_H$. Further let $\Phi : G \rightarrow H$ be a homomorphism. The kernel from $\Phi$ is the set

$$Kernel \Phi = \{g \in G |\Phi(g) = e_H\}$$


  1. Show that the kernel $\Phi$ is a subgroup from $G$.

  2. $\Phi$ is injective, when kernel $\Phi$ = $\{e_G\}$ is true.

What I have so far:

At number 1, isn't this true because the image from the function $\Phi$ from every element from $G$ is the neutral element? At number 2, if kernel $\Phi$ = $\{e_G\}$, then the homomorphism is true, this means $\Phi$ is injective.

b) Show that the set $G \times H$ with the operation $$(g_1, h_1)\star (g_2,h_2)=(g_1\bullet g_2, h_1*h_2)$$ is a group. Further, show that the function $$\Phi: (G \times H, *)\rightarrow (G, \bullet), (g,h)\rightarrow g,$$ ist a homomorphism. Define the kernel from $\Phi$.

I'm completely lost here. :(

Thanks A LOT in advance for any tips!

  • 3
    $\begingroup$ To show that $\Phi: \mathbb{Z} \rightarrow \mathbb{Q}$ is a homomorphism, you need to show that $\Phi(a+b) = \Phi(a)+\Phi(b)$ for any $a, b \in \mathbb{Z}$. That doesn't work for $\Phi(x) = x^2$ (for example, $a = b = 1$). $\endgroup$
    – Adam Saltz
    Nov 13, 2011 at 14:16

2 Answers 2


Note that $(\mathbb{Z},+)$ and $(\mathbb{Q},+)$ are groups under the binary operation "$+$". Now if $f: \mathbb{Z} \to \mathbb{Q}$ is a group homomorphism it has to satisfy :

  • $f(x+y)= f(x) + f(y)$ which clearly $f(x)=x^{2}$ doesn't satisfy because $(x+y)^{2} \neq x^{2}+y^{2}$.

As for your question $a)$ note that $\operatorname{Ker}(\Phi)$ is a subgroup of $G$ and is also a normal subgroup. This can be seen as follows:

  • Take $x,y \in \operatorname{Ker}(\Phi)$. Note that then we have $\Phi(x) = e_{H}$ and $\Phi(y)=e_{H}$. Now, $\Phi(xy)= \Phi(x) \ast \Phi(y) = e_{H} \ast e_{H} = e_{H}$ which says that $xy \in \operatorname{Ker}(\Phi)$. Similarly show that if $x \in \operatorname{Ker}(\Phi)$ then $x^{-1} \in \operatorname{Ker}(\Phi)$.

  • To show $\Phi$ is injective $\operatorname{Ker}(\Phi)={e_{H}}$, note that if $x,y \in \operatorname{Ker}(\Phi)$ and $\Phi(x)= \Phi(y) \Rightarrow \Phi(x \bullet y^{-1}) = e_{H} \Rightarrow x=y \Rightarrow \Phi$ is injective.

b) Is very trivial. Just check whether the set satisfies the four group axioms.

  • $\begingroup$ Hi! Thank you very much for the help! Can you help with the proof that $x^{-1}$ is in $\text{Ker}(\Phi)$? Here's what I have so far. $\Phi(x) = e_{H}$ and $\Phi(x*x^{-1}) = \Phi(x) \ast \Phi(x^{-1}) = e_H*e_H = e_{H}$. Is this correct? $\endgroup$
    – Clash
    Nov 13, 2011 at 16:35
  • $\begingroup$ @Clash: Yes, well done :) $\endgroup$
    – user9413
    Nov 13, 2011 at 16:38
  • $\begingroup$ really?! I had no faith in my answer!! god, I'm such a pro at math, lol, I suck $\endgroup$
    – Clash
    Nov 13, 2011 at 16:50
  • $\begingroup$ When you say "Take $x,y \in \operatorname{Ker}(\Phi)$", does it mean, for every $x,y$ or exists $x,y$? $\forall$ or $\exists$? $\endgroup$
    – Clash
    Nov 13, 2011 at 17:01

You can check here to see the definition of a homomorphism. The map $\phi$ is not a group homomorphism because it doesn't preserve the grouop operation, which is addition in $\mathbb{Z}$ and $\mathbb{Q}$. You can easily see that $\phi(2+3) = 5^2 = 25$, while $\phi(2)+\phi(3) =2^2 + 3^2 = 13$. They are not equal. We need to have $\phi(xy) =\phi(x)\phi(y)$ for $\phi$ to be a group homomorphism.

I am not an English speaker, and I don't understand when you say "the image from the function $\phi$ from every element from $G$ is the neutral element". In order to prove that the kernel of $\phi$ is a subgroup, you can check for any two elements $x$ and $y$ in the kernel, both $x^{-1}$ and $xy$ are in the kernel, i.e., show the conditions of a group is satistied. This is not difficult considering the definition of a homomorphism of groups. In fact, you can only show $x^{-1}y$ is in the kernel, because this is equivalent.

To show the injectivity of $\phi$ when the kernel is $\{ e_G \}$, consider two elements $x$ and $y$ such that $\phi(x) = \phi(y)$. We have $\phi(xy^{-1}) = \phi(x)\phi(y)^{-1} = e_H$, because $\phi$ is a homomorphism. Then $xy^{-1} = e_G$ because it's in the kernel. So, $x = y$, and $\phi$ is injective.

In order to show the product is a group, you also have to check the conditions of closedness under production and inverse. The map $\phi$ is a homomorphism of groups because it satisfies the definition, preseving the operations of production and inverse ($\phi(x^{-1}) = \phi(x)^{-1}$, $\phi(xy) = \phi(x)\phi(y)$).

I hope I have made everything clear. If I am not, please let me know :) I've been helped many times at this site, and I will be happy if this helps you.

  • $\begingroup$ Hi thank you very much for the enlightening examples! You have helped me very much! I'm still doing a), I will tell you when I get to b) if I got everything. Thanks again! $\endgroup$
    – Clash
    Nov 13, 2011 at 16:36

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