Prove the function is a homomorphism and calculate the image

Question

Consider the function $f: (Z_{10}, +_{10}) \to (Z_{20}, +_{20})$ defined by $f(n)=2n$. Show that $f$ is a homomorphism and calculate the image.

[NOTE: $Z_m = \{0,1,2,...,m-1\}$ is a group under the operation $+{_m}$ defined via $i +_m j$ equals the remainder of $i+j$ when divided by $m$.]

I understand that to prove homomorphism I need to show that the Homomorphic Property holds: $f(x*y)=f(x)*f(y)$. But I'm not entirely sure how to go about it using the $+_m$ notation. I also have no idea how to find the image, which is defined by $\{f(g)|g \in G\}$.

Answer 1

Let $x,y \in Z_{10}$ and say $x+y=z$ (within the additive group). Then, if we ignore the modulus and just consider the elements as integers, we have $x+y=z+t$ where $t$ is some multiple of $10$ (really, it'll just be either $0$ or $10$). Then $f(x)+f(y)=2(x+z)=2x+2y=2z+2t$, where this is without the modulus. With respect to the group, $f(z)=2z$. Can you see why these things, mod($20$), are equal?

EDIT: Also, with regard to the image, say $f(g)=h$ for $g \in Z_{10}, h \in Z_{20}$. What can you say is certainly true about $h$, no matter what $g$ is chosen?

Answer 2

The homomorphism property is just $f (x+y)=2 (x+y)=2x+2y=f (x)+f (y) $.

You can check that the kernel is $\mathbb Z_2\cong\{0,5\} $, and, by the first isomorphism theorem, the image is isomorphic to $\mathbb Z_{10}/\mathbb Z_2\cong\mathbb Z_5$. It is the subgroup $\{0,4,8,12,16\} $, the only $\mathbb Z_5$ in $\mathbb Z_{20} $.