Question: Let $\phi$ : $\mathbb Z \to \mathbb Z$ be given by $\phi (n)$=$7n$.
Prove that $\phi$ is a group homomorphism. Find the kernel and image of $\phi$

Definition of homomorphism: Let $G_1$ and $G_2$ be two groups. Then $\phi$ : $G_1 \to G_2$ is called a homomorphism iff $\forall$ $a,b \in G_1$. So $\phi (ab)$= $\phi (a)$$\phi (b)$

Definition of kernel: If $\phi$ : $G_1 \to G_2$ is a homomorphism
Then {$g \in G_1$ : $\phi (G_1)$=$e_2$} is called the kernel of $\phi$ denoted by ker $\phi$
Alternate definition of kernel of $\phi$ : Ker $\phi$ :={$x\in G_1 : \phi (x)$=$e_2$} note: $\phi$$^{-1}(e_2)$ is also included.

Here is my attempt:
$\phi (n)$=$7n$ for some $n\in \mathbb Z$
So $\phi$ is onto

$\phi(n_1)$+$\phi(n_2)$=$\phi(n_1 + n_2)$

LHS=RHS (lefthand side is equal to right hand side)
therefore, $\phi$ is $1$- $1$

I am struggling to find the kernel and image of $\phi$.
I need help. Also you can check the work I have done so far if I made some errors

  • $\begingroup$ Your proof for the fact that $\phi$ is a homomorphism is at least confusing. You have to prove that $\phi(n_1+n_2) = \phi(n_1) + \phi(n_2)$. $\endgroup$ – Friedrich Philipp Mar 18 '17 at 2:46
  • 1
    $\begingroup$ $\phi$ is not onto. Also, it seems from what you've written that you don't know what 1-1 and onto mean. $\endgroup$ – Omnomnomnom Mar 18 '17 at 2:48
  • $\begingroup$ @FriedrichPhilipp I did prove $\phi(n_1 + n_2)$=$\phi(n_1)$+$\phi(n_2)$. Hence, LHS=RHS $\endgroup$ – behold Mar 18 '17 at 2:54
  • 2
    $\begingroup$ There are already comments and answers regarding homomorphism. The kernel is given by $$Ker \phi=\{n\in\Bbb Z:\phi(n)=0\}=\{n\in\Bbb Z:7n=0\}=\{0\}$$ and $$Im \phi=\{f(n):n\in\Bbb Z\}=\{7n:n\in\Bbb Z\}=\{\dots,-14,-7,0,7,14,\dots\}=7\Bbb Z.$$ $\endgroup$ – Juniven Mar 18 '17 at 3:06
  • $\begingroup$ @behold: No,you did not! Why don'T you trust in more experienced people than yourself? $\endgroup$ – Friedrich Philipp Mar 18 '17 at 3:07

The homomorphism proof as it currently stands is incorrect.

You have:

\begin{align*} \phi(n_1)+\phi(n_2) & \color{red}{=} \phi(n_1+n_2) \\ & \color{red}{=} \phi(n_1)+\phi(n_2) \\ & = 7n_1+7n_2 \\ & = 7(n_1+n_2) \end{align*}

You have the important elements here, but they're arranged in a confusing way, that I would consider incorrect. Specifically, the red equalities above are incorrect.

If you have a group $(G,+)$, and another group $(H,\oplus)$, a group homomorphism $\phi:G\to H$ is a function $\phi$ such that, for all $x,y\in G$: $$\phi(x+y) = \phi(x)\oplus\phi(y)$$ Note that there are different operations "inside" $\phi$'s brackets vs outside of them. This is because the elements $x,y\in G$ (so are combined using $G$'s operation), but $\phi(x),\phi(y)\in H$ (so are combined using $H$'s operation).

A homomorphism is a function that satisfies the above. So, to prove something is a homomorphism, you need to prove that it always satisfies that equation.

Let $\phi:\mathbb Z\to\mathbb Z$ be defined by $x\mapsto 7x$. Then, we have that $\phi(x+y) = 7(x+y)$. We also have that $\phi(x)+\phi(y) = 7x+7y$. Can we show that these expressions are equal? If we can, then $\phi$ is a homomorphism.

Showing they're equal isn't too bad, what you do is say that: $$\phi(x+y) = 7(x+y) \color{red}{=} 7x+7y = \phi(x)+\phi(y)$$ Everything here I've written is just "writing what $\phi(x)$ means" besides the equality in red, which using distributivity of addition in $\mathbb Z$. But, what we've done here is start with $\phi(x+y)$, and show how that must be equal to $\phi(x)+\phi(y)$, so $\phi$ must be a homomorphism.

  • $\begingroup$ I see. I should have stopped after the LHS was done and did the RHS separately $\endgroup$ – behold Mar 18 '17 at 3:04

Your proof that $\phi$ is a homomorphism is correct. However, be careful. You said "$\phi$ is 1-1 or a homomorphism." These are not equivalent concepts. 1-1 means that $\phi(x) = \phi(y)$ implies $x = y$, i.e. $\phi$ never maps two different elements to the same image.

The kernel is the set $n \in \mathbb{Z}$ with $\phi(n) = 0$. If $7n = 0$, then what is $n$?

The image is the set of all images. What integers can be written in the form $7n$?

  • 2
    $\begingroup$ OP also wrote that $\phi$ is onto (i.e., surjective) which is definitely not the case. $\endgroup$ – Friedrich Philipp Mar 18 '17 at 2:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.