Question

find ker(ϕ) and ϕ(18) for ϕ:Z→Z10 such that ϕ(1)=6

Answer 1

the answer given :
ker(ϕ)=5Z because 6 has order 5 in Z10
why we have to consider the order of 6?

Answer 2

Could you specify what you mean by Z and Z10? Do you mean the following?\[\mathbb Z = \{\ldots,-2,-3,0,1,2,3,\ldots\}\]\[\mathbb Z_{10}=\{1,2,3,\ldots,10\}\]

Answer 3

yup...but \[Z _{10}={0,1,2,.....,9}\]

Answer 4

Can I have a little more context? Is this for a class on Group Theory?

Answer 5

yup...

Answer 6

\[\phi(5)=\phi(1+1+1+1+1)=5 \phi(1)=0\]
for ϕ(-5)?

Answer 7

how we get \[\phi(-5)\]=0?

Answer 8

Let's tackle the first question. The kernel of \(\phi\) is defined as the set of points in \(\mathbb Z\) such that when we apply the group homomorphism \(\phi\), we get the identity of \(\mathbb Z_{10}\), which is simply 0, if we are considering modulo 10 addition.\[\text{ker}(\phi)\overset{\text{def}}{=} \left\{ z \in \mathbb Z : \phi(z) = 0\right\}\]Since we are given \(\phi(1)=6\), we can expand \(\phi(z)\) as follows. Here, I denote \({+}_{10}\) as addition modulo 10.\[\large \phi(z)=\phi(\underbrace{1+1+\cdots+1}_{\text{$z$ times}}) = \underbrace{\phi(1) {+}_{10}\phi(1) {+}_{10}\cdots {+}_{10} \phi(1)}_{\text{$z$ times}} \]\[\large = \underbrace{6 {+}_{10} 6 {+}_{10}\cdots {+}_{10} 6}_{\text{$z$ times}} \]It's clear that this is true for \(z=0,5,10,15,\ldots\), and we can generalize that to all multiples of five, including negative multiples. To be honest, I'm not 100% sure how you'd prove that -5 is in the kernel just given the given information, though you can reason it out just by seeing the pattern of multiples of fives. I think this gives you enough information to evaluate at \(z=18\), too.

Answer 9

thank you for your explanation...i really need that..hehe..thanks ya...