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Compute $\ker(\phi)$ for $\phi:\mathbb{Z}\to\mathbb{Z}_7$ such that $\phi(1)=4$

My answer say thatenter image description here

I have no idea what they mean by "4 has order 7 in $\mathbb{Z}_7$"

All I can write is that $\ker(\phi) = \{ m \in \mathbb{Z} : \phi(m) =0\}$ because $0$ is the identity in $\mathbb{Z_7}$ right? But I donn't know what $\phi$ is, so I don't know how to continue. And I have no idea why theya re breaking up the sum inside $\phi$

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up vote 3 down vote accepted

You know that $\varphi$ is a homomorphism, and you know that $\varphi(1)=4$. Therefore

$$\begin{align*} \varphi(2)&=\varphi(1+1)=\varphi(1)+_7\varphi(1)=4+_74=1\\ \varphi(3)&=\varphi(2+1)=\varphi(2)+_7\varphi(1)=1+_74=5\\ \varphi(4)&=\varphi(3+1)=\varphi(3)+_7\varphi(1)=5+_74=2\\ \varphi(5)&=\varphi(4+1)=\varphi(4)+_7\varphi(1)=2+_74=6\\ \varphi(6)&=\varphi(5+1)=\varphi(5)+_7\varphi(1)=6+_74=3\\ \varphi(7)&=\varphi(6+1)=\varphi(6)+_7\varphi(1)=3+_74=0\;,\text{ and}\\ \varphi(8)&=\varphi(7+1)=\varphi(7)+_7\varphi(1)=0+_74=4\;. \end{align*}$$

Here I’m using the homomorphism property: $\varphi(m+n)=\varphi(m)+_7\varphi(n)$ for all $m,n\in\Bbb Z$.

Clearly the values of $\varphi(n)$ will cycle through the pattern $4,1,5,2,6,3,0$ repeatedly. The length of the cycle is $7$, so every seventh value of $\varphi(n)$ will be $0$, starting with $\varphi(7)$; from that it’s not hard to see that

$$\ker\varphi=\{n\in\Bbb Z:\varphi(n)=0\}=\{n\in\Bbb Z:7\mid n\}=7\Bbb Z\;,$$

the set of multiples of $7$.

To say that $4$ has order $7$ in $\Bbb Z_7$ just means that the smallest positive integer $n$ such that $$\underbrace{4+_7\ldots+_74}_n=0\text{ in }Z_7$$ is $n=7$. We saw this in the chart above: starting with $4$ and repeatedly adding $4$ produced in turn $4,1,5,2,6,3,0$, the cycle that we already noted, and it wasn’t until we’d added together seven $4$’s that we got the additive identity $0$ of $\Bbb Z_7$.

They were using the fact that $\varphi$ is homomorphism when they calculated $\varphi(25)$. First, $25=21+4$, so by the homomorphorphism property $\varphi(25)=\varphi(21)+_7\varphi(4)$. Now $21$ is a multiple of $7$, so $21\in\ker\varphi$, and $\varphi(21)=0$, and therefore $\varphi(25)=0+_7\varphi(4)=\varphi(4)$. Then they split $4$ as $1+1+1+1$ and used the homomorphism property again:

$$\begin{align*} \varphi(4)&=\varphi(1)+_7\varphi(1)+_7\varphi(1)+_7\varphi(1)\\ &=4+_74+_74+_74\\ &=(4+_74)+_7(4+_74)\\ &=1+_71\\ &=2\;. \end{align*}$$

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so how did they know the order so quickly without doing the brute force? I mean even $\underbrace{4+_7\ldots+_74}_n=0\text{ in }Z_7$ for me takes a while to compute – Hawk Nov 18 '12 at 21:03
@sizz: Because $7$ is prime, so every every non-zero element of $\Bbb Z_7$ has order $7$. Lagrange’s theorem: the order of an element must divide the order of the group. The only divisors of $7$ are $1$ and $7$, so every element that isn’t the identity must have order $7$. – Brian M. Scott Nov 18 '12 at 21:06
If I were to tweak the problem a little. What would happen if we were to map $\mathbb{Z} \to \mathbb{Z_8}$?. The divisors are 4 and 2. After working out the algebra, it turns out the order of $4$ is $2$ in $\mathbb{Z}_8$. So by Lagrange's theorem, Is it just |8|/|4| = 2? Also I thought Lagrange Theorem say that the order of a subgroup must divide the order of the group, why do you say 'element'? – Hawk Nov 18 '12 at 21:19
I also have another example where $\phi: \mathbb{Z} \to \mathbb{Z_10}$ where $\phi(1) = 6$ and the book again says it is $5$. I had to work out the brute force method to verify and using Lagrange's theorem too I got 2 and 5 as my divisor for 10. But I didn't know which to pick – Hawk Nov 18 '12 at 21:20
@sizz: (1) If $\varphi:\Bbb Z\to\Bbb Z_8$ and $\varphi(1)=4$, then $\ker\varphi=2\Bbb Z$: $\varphi$ takes every even integer to $0$ in $\Bbb Z_8$. (2) The order of an element $x$ of a group $G$ is equal to the order of the subgroup $\langle x\rangle$ of $G$ generated by $x$, so Lagrange’s theorem does give the result that I stated. (3) In $\Bbb Z_{10}$ the elements of order $5$ are the multiples of $2$, $5$ is the only element of order $2$, and the other odd numbers have order $10$. In $\Bbb Z_n$ the order of an element $k$ is $\dfrac{n}{\gcd(k,n)}$. – Brian M. Scott Nov 18 '12 at 21:31

Well, for sure $\,7\Bbb Z\subset \ker\phi\,$ , right? Because

$$\phi(7k):=7k\phi(1)=28k=0\pmod 7$$


$$n\in\ker\phi\Longrightarrow 4n=0\pmod 7\Longrightarrow n=0\pmod 7\,\,(\text{since}\,\,(4,7)=1)\Longrightarrow$$

$$\Longrightarrow \ker\phi\subset 7\Bbb Z$$

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Is there anyway to dumb it down even further for me? – Hawk Nov 18 '12 at 20:06

Start computing. Because $\phi(1)=4$, it follows that $\phi(2)=\phi(1+1)=4+4=1$. (Remember, we are working in $\mathbb{Z}_7$.) Keep calculating. We get $\phi(3)=\phi(2+1)=1+4=5$, $\phi(4)=2$, $\phi(5)=6$, $\phi(6)=3$, and $\phi(7)=0$. Bingo!

It is now fairly easy to see that $\phi(n)=0$ precisely if $n$ is a multiple, positive or negative, of $7$. For any integer $n$ can be expressed as $n=7q+r$, where $0\le r\le 6$. So now we know the kernel of $\phi$.

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"4 has order 7" means that $$4+4+4+4+4+4+4$$ (7 times) is the identity element in $\Bbb Z_7$, and no smaller (non-trivial) sum of fours is.

For any homomorphism $\psi:G\to H$ between groups $G$ and $H$ (for notational simplicity I'm assuming they are abelian, although the exact analogus result holds for non-abelian groups) you have that $$ \psi(a+_Gb) = \psi(a)+_H\psi(b). $$ This means that if we know $a$ to be in the kernel of $\psi$, then $\psi(a+_G b) = 0_H +_H\psi(b) = \psi(b)$. It also means that if $G=\Bbb Z$, then $\psi$ is completely determined by the value of $\psi(1)$. In your case, $$\phi(4) = \phi(1+1+1+1) = \phi(1) +_7\phi(1) +_7\phi(1) +_7\phi(1) = 2_7.$$This ability use the homomorphism on each term in turn is the reason they break it up in your answer.

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