Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\phi$

share|cite|improve this question
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:

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

share|cite|improve this answer
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

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


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

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

share|cite|improve this answer
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 $\phi$.

share|cite|improve this answer

"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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.