-1

Kindly vet and guide:

First of all can we consider a cyclic group as both : multiplicative and additive group.

Say, $C_n$ has at least two generators $g, g^{n-1}$ and though composition adds exponents, but group operation is multiplication.

To take as additive group, need consider the fact that $C_n\cong \mathbb{Z_n}.$

Request reference source text, as for this have relied on lines stated as part of answer here only.

Second, if can consider as multiplicative group; then how to use Bezout lemma to find inverse, as shown here.

Say, find Inverse of $3$ in multiplicative group $C_{20}.$

$3x + 20y= 1,$ Have a doubt that $x,y \in \mathbb {Z}$ rather than $\in \mathbb {Z} / 20\mathbb{Z}.$

$$\begin{align}&20= 3\cdot 6+2\\& 3=2\cdot 1+1\end{align}$$

Now, taking reverse, have:

$$1= 3-2\cdot 1,$$ $$1= 3-(20-3\cdot 6)\cdot 1$$ But, unable to proceed.

Sourav Ghosh
  • 12,997
jiten
  • 4,524
  • 2
    Bezout's theorem is not applicable in this case, since it is a statement over $\mathbb{Z}$ rather than $\mathbb{Z}/n\mathbb{Z}$. It is probably more straightforward to do this problem by inspection. – JD1874 Aug 11 '22 at 10:07
  • @m1820 Any reference source please. That would be very helpful in quoting. Though, what you said seems obvious, but still if a text were available to state this; then best. Might be can learn more from that text. – jiten Aug 11 '22 at 10:13
  • @SouravGhosh The generator $g$ of $C_n$ has subgroup $\langle g\rangle={g, g^2, g^3,\cdots , g^{n-1}}.$ The equivalent subgroup , under additive operation, is:$\langle g\rangle={ g, 2g, 3g, 4g, \cdots, (n-1)g}.$ – jiten Aug 11 '22 at 10:25
  • Most all the known methods for computing modular inverses (and solving linear congruences) have already been discussed here at length many times, e.g. see the linked dupes (and their links), e.g. $\bmod 20!:\ \dfrac{1}{3}\equiv \dfrac{1}3\dfrac{\color{#c00}7}{\color{#c00}7}\equiv \dfrac{7}1,$ where $\color{#c00}7$ is the first multiple of $,3,$ that's $> 20 = $ modulus (cf. Gauss's algorithm) – Bill Dubuque Aug 11 '22 at 14:46
  • The hom $,n\to g^n,$ from $,\Bbb Z$ to $,C_n,$ has kernel $20\Bbb Z$ so $,C_n\cong \Bbb Z/20,$ by the first isomorphism theorem. Please ask only one question per post. – Bill Dubuque Aug 11 '22 at 14:56
  • 1
    You can solve $3x+20y=1$ using the Extended Euclidean Algorithm. One method is outlined in this answer. Once you have $3x+20y=1$, you have $x\equiv3^{-1}\pmod{20}$. Furthermore, $3x+20y=1$ also says that $(x,20)=1$ so $x$ must also be in the multiplicative group mod $20$. – robjohn Aug 11 '22 at 15:27
  • @rob Kudos - it surely helps having diamond mods set good examples for site policies. – Bill Dubuque Aug 11 '22 at 15:38
  • @BillDubuque: now I'm waiting for the person who will flag my comment for answering in comments. – robjohn Aug 11 '22 at 15:48
  • @Rob Ha, ha, but that's really a problem with the question (more than one in one post) rather than the answer(er). – Bill Dubuque Aug 11 '22 at 16:19
  • @BillDubuque According to your logic this two post must be duplicate of one another. First one weierstrass approximation theorem and second one separability of the space $(C[0, 1], |•|_{\infty}) $ . There are more... https://math.stackexchange.com/q/1161111/977780 https://math.stackexchange.com/a/644304/977780 – Sourav Ghosh Aug 11 '22 at 17:12
  • @Sourav I have no idea what you are talking about. But see abstract duplicates. – Bill Dubuque Aug 11 '22 at 17:16
  • @BillDubuque Weierstrass approximation theorem is basically density of polynomial in the space of continuous functions. Hence two questions are same. By the way, you are doing a great job. I really appreciate . – Sourav Ghosh Aug 11 '22 at 17:19
  • 1
    @Sourav I thought you were implying that it had something to do with this topic. Of course there are very many other (abstract) duplicates that have not yet been organized. Alas, we are far understaffed when it comes to site organization (common for gamified platforms). Thanks for your support. – Bill Dubuque Aug 11 '22 at 18:09
  • @robjohn In $3x+20y=1 $you stated $x\in \varphi(20)=8$ with members: ${1,3,5,7,11,13,17,19}. $ Will the same not apply to $y $ variable too, i.e.$(y,3)=1 $ and $y\in \varphi(3)=2$ with set $={1,2}$? Also any reference please. Also, if for this simple equation could be shown a solution. – jiten Aug 11 '22 at 18:54

1 Answers1

1

I am going to answer of your original question.

Let $U(20) $ be the multiplicative group of $C_{20}$.

We have $3\in U(20)$. What is the inverse of $3$ in $U(20) $?

Suppose $3^{-1}=x$. Then $3x\equiv 1\pmod{20}$. We have $1=3-(20-3\cdot 6)=3\cdot 7-20$. Hence $3\cdot 7\equiv 1\pmod{20}$. Hence $x=7$ is the required inverse of $3$. You can recheck your work: $3\cdot 7\pmod{20}=21\pmod{20} =1$.

Another approach: $(3, 20) =1$ . Hence by F.L.T $3^{\varphi(20)} \equiv 1\pmod{20}$, which implies $3^8\equiv 1 \pmod{20}.$ But

$$\begin{align}3^{-1}&\equiv 3^{-1}\cdot 3^8 \pmod{20}\\ &\equiv 3^7\pmod{20} \\ & \equiv3^43^3 \pmod{20}\\ & \equiv 7\pmod{20}. \end{align}$$

Shaun
  • 44,997
Sourav Ghosh
  • 12,997
  • Have a request: find element $a$ in $\mathbb {Z_{180}}$ s.t. $a^{60}=e=1.$ How to address it. Or, if post a seperate question, then need state some work on it. – jiten Aug 11 '22 at 11:33
  • 1
    $|U(180) |=\varphi(180) =48$ . Order of an element divides order of the group. Hence $|a|=60$ isn't possible. But $a^m=e$ implies $|a|\mid m$ . Hence you can choose $a$ such that $|a|\mid 60$ and $|a|\mid 48$ . More specifically choose $a$ such that $|a|$ is the common divisor of $ 60$ and $48$. – Sourav Ghosh Aug 11 '22 at 11:40
  • So, the condition that $|U(n)|=\varphi(n)$ decides number of elements in a multiplicative group applies to all cyclic groups, due to need to have multiplicative inverse. Also, the common divisor of $60, 48$ are:${1,2,3,4,12}.$ The question is having another element $b$ with $|b|= 36.$ Hence, common factors of $36,48={1,2,3,4,6}.$ Next, asks to find possible order values of $ab.$ As $|a|\ne |b|,$ so $a\ne b.$ And $ab\ne 1$ too. To find possible orders then need all possible $(a,b)= (1,2), (1,3), (1,4), (1,6), \cdots,(12,1),(12,3),(12,4),(12,6)}.$ Then, find LCM of each pair. – jiten Aug 11 '22 at 12:22
  • 1
    $ab$ and $(a, b) $ are two different things. For the latter case one need direct product. Well if you want to know more, you should start a separate thread. – Sourav Ghosh Aug 11 '22 at 12:25
  • But, if $|a|\ne 60,$ then multiplicative group is not possible. Hence, either the question is wrong; or it is an additive group. For framing new post with emphasis on product groups, will post soon. – jiten Aug 11 '22 at 12:46
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Aug 11 '22 at 14:44
  • @BillDubuque It's my mistake. It's not the exact dupe but closed to a dupe or even can be found using the dupe. Will try not to repeat such mistake. – Sourav Ghosh Aug 11 '22 at 17:00
  • @Sourav Of course there any many tens (if not hundreds) of prior posts mentioning that well known method, e.g. here and here and here. – Bill Dubuque Aug 11 '22 at 17:15