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I have come up with the following question/problem and want to try to understand and solve it.

Basically, given a cyclic group $G = \mathbb{Z}_{59}^{\times}$; find inverse of $31$ in this group.

Now, I understand that $\mathbb{Z}^{\times}$ in a cyclic group implies that it is multiplicative but then how do I expand the cyclic group? More importantly, how do I find inverses of a number in a group say $31$?

Con
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user23473433819233d
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1 Answers1

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We have that $\text{gcd}(31,59) = 1$. Now use the extended euclidean algorithm to find the Bezout coefficients to get $1 = 31a + 59b$ for some integers $a$ and $b$. This means that you get $1 = 31 \cdot a$ mod $59$, i.e. $a$ mod $59$ is the inverse of $31$ mod $59$.

As you asked for the computation:

We have

$$59 = 1 \cdot 31 + 28$$

$$31 = 1 \cdot 28 + 3$$

$$28 = 9 \cdot 3 + 1$$

$$3 = 1 \cdot 3 + 0,$$

which once again shows that $\text{gcd}(59,31) = 1$. We now get

$$\small 1 = 28 - 9 \cdot 3 = 28 - 9(31- 28) = -9 \cdot 31 + 10 \cdot 28 = -9 \cdot 31 + 10(59 - 31) = 10 \cdot 59 -19 \cdot 31,$$ i.e. $a = -19$ and $b = 10$. Thus $\overline{-19} = \overline{40}$ is the inverse of $\overline{31}$.

Let me also add a concrete description of the group. Consider $\mathbb{Z}/n\mathbb{Z} = \lbrace \overline{0},\overline{1},\dots,\overline{n-1} \rbrace$. Then we get $\mathbb{Z}/n\mathbb{Z}^{\times} = \lbrace \overline{k} \mid \text{gcd}(k,n) = 1\rbrace$. In the case that $n$ is prime (as in your example) this yields $\mathbb{Z}/n\mathbb{Z}^{\times} = \lbrace \overline{1},\dots,\overline{n-1} \rbrace = (\mathbb{Z}/n\mathbb{Z})\setminus\lbrace \overline{0} \rbrace$.

Con
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  • Thanks for providing an answer. Just a few questions, why did you jump/do the gcd(31,59)? Also how do you know that the gcd is 1? I assume it is because they are both prime numbers right? – user23473433819233d Aug 13 '19 at 00:30
  • Also, how do I expand the group? It is all interesting stuff but also complex to understand which is why I was hoping that an example of my own making would help me better understand cyclic groups in general. Would definitely make my day if I understood it!!! – user23473433819233d Aug 13 '19 at 00:32
  • The extended euclidean algorithm allows you to get such an equation for the gcd. Therefore it was important that the gcd is $1$. Yes. The gcd is $1$ since $31$ (actually also $59$) is a prime number. One of them being prime is enough though as $31 \neq 59$. – Con Aug 13 '19 at 00:33
  • What do you mean by ”expand the group“? Do you want a concrete description of it? – Con Aug 13 '19 at 00:35
  • I see, so what you are saying is that so long as one of the numbers are prime then the gcd is 1? I don't suppose you could kindly help with expanding the cyclic group? – user23473433819233d Aug 13 '19 at 00:35
  • Usually Z*59 = <1,2,3,....> – user23473433819233d Aug 13 '19 at 00:36
  • Oops! There was a typo above. I meant to say that $31$ does not divide $59$. A prime number has two divisors: Itself and $1$ And we were able to exclude itself in that case which leads to $1$ being the only common divisor. Let me add that to my answer. I will edit it. – Con Aug 13 '19 at 00:40
  • Thanks for your time Sir, are you able to apply that example to the cyclic group please? I think I'm starting to get this but it would be great to see the theory being applied practically to my own problem – user23473433819233d Aug 13 '19 at 00:51
  • Well... I just did. You only need to substitute $n$ for $59$ in the latter equation. This means that the group that you are considering consists of $1$ mod $59$, $2$ mod $59$, ..., $58$ mod $59$. I am glad to help more, but you will need to tell me more precisely what you are asking. That is not clear to me. – Con Aug 13 '19 at 00:59
  • No problem Sir, so just a few questions to clarify. – user23473433819233d Aug 13 '19 at 01:01
  • So Z*59 = ,<1,2,3...,58>?
    • – user23473433819233d Aug 13 '19 at 01:02
  • I'm stuck and clueless on how to use the extended euclidean algorithm for bezout in my case. i watch a youtube video but im lost because in our case it has to equal 1. Surely a=0 and b=0?
    • – user23473433819233d Aug 13 '19 at 01:03
  • As in, it would be ever so kind and grateful if you could kindly type out worked example to find the inverse in this case, if it is not too much to ask? If not, no worries – user23473433819233d Aug 13 '19 at 01:04
  • You are asking whether the group is generated by all its elements, but a group is always generated by all its elements, which is why I believe that you want to say something else. If you want to list its elements you have to use $\lbrace$ and $\rbrace$ and not $\langle$ and $\rangle$.
    • – Con Aug 13 '19 at 01:05