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
Best Answer
The homomorphism proof as it currently stands is incorrect.
You have:
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.