Homomorphism from $\Bbb{Z}_{10}$ to $\Bbb{Z}_5$

Question

Let $f$ : $\Bbb{Z}_{10}$ $\rightarrow$ $\Bbb{Z}_5$ be defined as

For $x \in \mathbb Z_{10}, f(x) = 2x\pmod 5\in \mathbb Z_{5}$

I need to show that $f$ is a homomorphism.

Now here $f(\bar1) = \bar2$.

If $f$ is a homomorphism then order of $\bar2$ that is $5$ must divide order of $\bar1$ that is $10$ which is the case here.

I also know that $\bar1$ is a generator of $\Bbb{Z}_{10}$.

But how could this ensure that $f$ will definitely be a homomorphism?

Answer 1

An (additive) group homomorphism is a function $f : G \to H$ satisfying $f(a + b) = f(a) + f(b)$.

You can show that the map $f(x) = 2x\pmod 5\in \mathbb Z_{5}$ is a group homomorphism by show that for any two elements $a,b \in \mathbb Z_{10}$ that $f(a + b) = f(a) + f(b)$. Let's calculate both sides of that individually:

• $f(a + b) = 2 (a + b)$
• $f(a) + f(b) = 2 a + 2 b$

and these two formulas are equal by distributivity of multiplication. So we have a group homomorphism.

Answer 2

The map $f$ is the composition of multiplication by $2$ in $\mathbf Z/10\mathbf Z$, which is obviously an endomorphism of $\mathbf Z/10\mathbf Z$, by the canonical projection $\mathbf Z/10\mathbf Z\longrightarrow \mathbf Z/5\mathbf Z$ which results from the 3rd isomorphism theorem: $$\mathbf Z/10\mathbf Z\longrightarrow \mathbf Z/10\mathbf Z\big/5\mathbf Z/10\mathbf Z\simeq\mathbf Z/5\mathbf Z.$$