# Proving that $\phi$ is a group homomorphism. Also find the kernel and image of $\phi$

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

$\phi(n_1)$+$\phi(n_2)$=$\phi(n_1 + n_2)$
$\qquad\qquad\quad$=$\phi(n_1)$+$\phi(n_2)$
$\qquad\qquad\quad$=$7n_1$+$7n_2$
$\qquad\qquad\quad$=$7(n_1+n_2)$

$\phi(n_1+n_2)$=$7(n_1+n_2)$
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

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

• I see. I should have stopped after the LHS was done and did the RHS separately – 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 $7n$?

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