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)$$\phi (b)$Definition of kernel: If $\phi$ : $G_1 \to G_2$ is a homomorphismThen {$g \in G_1$ : $\phi (G_1)$=$e_2$} is called the kernel of $\phi$ denoted by ker $\phi$Alternate definition of kernel of $\phi$ : Ker $\phi$ :={$x\in G_1 : \phi (x)$=$e_2$} note: $\phi$$^{-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
 A: 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.
A: 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 $7n$?
