{"id":591,"date":"2020-08-09T06:24:02","date_gmt":"2020-08-09T06:24:02","guid":{"rendered":"http:\/\/soulofmathematics.com\/?page_id=591"},"modified":"2020-09-22T10:10:00","modified_gmt":"2020-09-22T04:40:00","slug":"cyclic-groups","status":"publish","type":"page","link":"https:\/\/soulofmathematics.com\/index.php\/group-theory\/cyclic-groups\/","title":{"rendered":"CYCLIC GROUPS"},"content":{"rendered":"\n

A group (G, \u00b7, e) is called cyclic if it is generated by a single element g. That is
if every element of G is equal to<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Note that if the operation is ‘+’, instead of using exponential we would use ng = g + g + g + ……<\/p>\n\n\n\n

FORMALLY STATING<\/em><\/strong><\/p>\n\n\n\n

Definition.<\/strong> A group \ud835\udc3a is a cyclic group if<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

In this case we say that G is a cyclic group generated by ‘a’, and obviously its an Abelian Group.<\/p>\n\n\n\n

Example.<\/strong> The set \u2124\ud835\udc5b = {0,1, \u2026 , \ud835\udc5b \u2212 1}(\ud835\udc5b \u2265 1) under addition modulo \ud835\udc5b is a cyclic group. Again, 1 and \u22121 (= \ud835\udc5b \u2212 1) are generators of \u2124\ud835\udc5b. It is worthwhile to note here that while the set of integers \u2124 has only two generators 1 and \u22121, \u2124\ud835\udc5b depending on the value of \ud835\udc5b may have more generators apart from 1 and \u22121. For example,\u212410 has 1, 3, 7, 9 (= \u22121) as its generators and \u212412 has 1, 5, 7, 11 (= \u22121) as its generators.<\/p>\n\n\n\n

Example.<\/strong> Consider the group \ud835\udc48(\ud835\udc5b)under multiplication modulo \ud835\udc5b, where \ud835\udc48(\ud835\udc5b) = {\ud835\udc58 \u2208 \u2115 \u2236 \ud835\udc58 < \ud835\udc5b and g.c.d. (\ud835\udc58, \ud835\udc5b) = 1} .<\/p>\n\n\n\n

Now for \ud835\udc5b = 10, \ud835\udc48(10) = {1, 3, 7, 9} = {30, 31, 33, 32} = \u23293\u232a. Also, \ud835\udc48(10) = {1,3,7,9} = {70, 73, 7, 72} = \u23297\u232a. Thus both 3 and 7 are generators of \ud835\udc48(10). Hence \ud835\udc48(10) is a cyclic group. Now we will show that \ud835\udc48(8) = {1, 3, 5, 7} is not a cyclic group. To show it we will find subgroup generated by each of the elements in \ud835\udc48(8). Observe that<\/p>\n\n\n\n

\u23291\u232a = {1}
\u23293\u232a = {1, 3} = {30, 31}
\u23295\u232a = {1, 5} = {50, 51}
\u23297\u232a = {1, 7} = {70, 71}<\/p>\n\n\n\n

Therefore, \ud835\udc48(8) \u2260 \u2329\ud835\udc4e\u232a for any \ud835\udc4e \u2208 \ud835\udc48(8) and hence the claim. Thus we have seen that \ud835\udc48(\ud835\udc5b) is a cyclic group or not depends on the choice of \ud835\udc5b.<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

GENERATORS OF A CYCLIC GROUP<\/h3>\n\n\n\n

Theorem 1.<\/strong> For any element \ud835\udc4e in a group \ud835\udc3a, \u2329\ud835\udc4e\u22121\u232a = \u2329\ud835\udc4e\u232a .In particular, if an element \ud835\udc4e is a generator of a cyclic group then \ud835\udc4e\u22121 is also a generator of that group.<\/p>\n\n\n\n

Theorem 2.<\/strong> For any element \ud835\udc4e in a group \ud835\udc3a, following holds:<\/p>\n\n\n\n

  1. If order of \ud835\udc4e is infinite, then all distinct powers of \ud835\udc4e are distinct elements i.e., \ud835\udc4e\ud835\udc56 \u2260 \ud835\udc4e\ud835\udc57 whenever \ud835\udc56 \u2260 \ud835\udc57, \ud835\udc56,\ud835\udc57 \u2208 \u2124.<\/li>
  2. If order of \ud835\udc4e is \ud835\udc5b for some \ud835\udc5b \u2208 \u2115, then \u2329\ud835\udc4e\u232a = {\ud835\udc52, \ud835\udc4e, \ud835\udc4e2, \u2026 , \ud835\udc4e\ud835\udc5b\u22121} and \ud835\udc4e\ud835\udc56 = \ud835\udc4e\ud835\udc57 \u21d4 \ud835\udc5b divides \ud835\udc56 \u2212 \ud835\udc57.<\/li><\/ol>\n\n\n\n

    Corollary.<\/strong> Let \ud835\udc3a be any group and \ud835\udc4e \u2208 \ud835\udc3a be an element of finite order \ud835\udc5b. If \ud835\udc4e^\ud835\udc58 = \ud835\udc52 for some \ud835\udc58 \u2208 \u2124, then \ud835\udc5b divides \ud835\udc58.<\/p>\n\n\n\n

    Theorem 3.<\/strong> Every group of prime order is cyclic and every element other than identity is a generator of the group.<\/p>\n\n\n\n

    GENERATORS OF FINITE CYCLIC GROUP<\/strong> <\/p>\n\n\n\n

    Let\ud835\udc3a = \u2329\ud835\udc4e\u232a be a finite cyclic. Then \ud835\udc3a = \u2329\ud835\udc4e\ud835\udc58\u232a \u21d4 \ud835\udc54.\ud835\udc50.\ud835\udc51 (\ud835\udc58, \ud835\udc5b) = 1 , where|\ud835\udc3a| = \ud835\udc5b .<\/p>\n\n\n\n

    Proof<\/strong>– Let \ud835\udc54.\ud835\udc50.\ud835\udc51 (\ud835\udc58, \ud835\udc5b) = 1, then there exist integers \ud835\udc60,\ud835\udc61 \u2208 \u2124 such that \ud835\udc58 + \ud835\udc5b = 1. But then \ud835\udc4e = \ud835\udc4e1 = \ud835\udc4e\ud835\udc58+\ud835\udc5b = \ud835\udc4e\ud835\udc58\ud835\udc4e\ud835\udc5b = (\ud835\udc4e\ud835\udc58)\ud835\udc60(\ud835\udc4e\ud835\udc5b)\ud835\udc61 = (\ud835\udc4e\ud835\udc58)\ud835\udc60 \u2208 \u2329\ud835\udc4e\ud835\udc58\u232a. Thus \ud835\udc4e \u2208 \u2329\ud835\udc4e\ud835\udc58\u232a, which further implies that all the powers of \ud835\udc4e belongs to \u2329\ud835\udc4e\ud835\udc58\u232a i.e., every element of \ud835\udc3a is in \u2329\ud835\udc4e\ud835\udc58\u232a. Hence \ud835\udc3a = \u2329\ud835\udc4e\ud835\udc58\u232a.<\/p>\n\n\n\n

    If \ud835\udc82 is a generator of a finite cyclic group, then \ud835\udc82\ud835\udc8c is a generator of the group if and only if \ud835\udc8c is relatively prime to order of \ud835\udc82. Thus a finite cyclic group of order \ud835\udc8f has \ud835\udf53(\ud835\udc8f) generators, where \ud835\udf53(\ud835\udc8f) is the number of positive integers less than \ud835\udc8f and relatively prime to \ud835\udc8f.<\/p>\n\n\n\n

    Here \ud835\udf53(\ud835\udc8f) is called Euler’s Totient Function<\/strong>.<\/p>\n\n\n\n

    In number theory, Euler’s totient function<\/strong> counts the positive integers up to a given integer n that are relatively prime to n. It is written using the Greek letter phi as \u03c6(n) or \u03d5(n), and may also be called Euler’s phi function<\/strong>. In other words, it is the number of integers k in the range 1 \u2264 k \u2264 n for which the g.c.d.(n, k) is equal to 1. The integers k of this form are sometimes referred to as totatives of n.<\/em><\/p>\n\n\n\n

    Example.<\/strong> The totatives of n<\/em> = 9 are the six numbers 1, 2, 4, 5, 7 and 8. They are all relatively prime to 9, but the other three numbers in this range, 3, 6, and 9 are not, since g.c.d.(9, 3) = g.c.d.(9, 6) = 3 and g.c.d.(9, 9) = 9. Therefore, \u03c6<\/em>(9) = 6. As another example, \u03c6<\/em>(1) = 1 since for n<\/em> = 1 the only integer in the range from 1 to n<\/em> is 1 itself, and g.c.d.(1, 1) = 1.<\/p>\n\n\n\n

    \"The
    The first thousand values of \u03c6<\/em>(n<\/em>). The points on the top line represent \u03c6<\/em>(p<\/em>) when p<\/em> is a prime number, which is p<\/em> \u2212 1.<\/figcaption><\/figure>\n\n\n\n
    \n
    IMAGE COURTESY<\/a><\/div>\n<\/div>\n\n\n\n

    Euler’s product formula<\/strong><\/p>\n\n\n\n

    It states<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    where the product is over the distinct prime numbers dividing n<\/em>.<\/p>\n\n\n\n

    The proof of Euler’s product formula depends on two important facts.<\/p>\n\n\n\n

    The function is multiplicative<\/strong><\/p>\n\n\n\n

    This means that if g.c.d.(m<\/em>, n<\/em>) = 1, then \u03c6<\/em>(mn<\/em>) = \u03c6<\/em>(m<\/em>) \u03c6<\/em>(n<\/em>). (Outline of proof<\/em>: let A<\/em>, B<\/em>, C<\/em> be the sets of non-negative integers, which are, respectively, coprime to and less than m<\/em>, n<\/em>, and mn<\/em>; then there is a bijection between A<\/em> \u00d7 B<\/em> and C<\/em>, by the Chinese remainder theorem<\/a>.)Value for a prime power argument.<\/p>\n\n\n\n

    If p<\/em> is prime and k<\/em> \u2265 1, then<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    Proof<\/em>: since p<\/em> is a prime number the only possible values of gcd(p<\/em>k<\/em><\/sup>, m<\/em>) are 1, p<\/em>, p<\/em>2<\/sup>, …, p<\/em>k<\/em><\/sup>, and the only way for gcd(p<\/em>k<\/em><\/sup>, m<\/em>) to not equal 1 is for m<\/em> to be a multiple of p<\/em>. The multiples of p<\/em> that are less than or equal to p<\/em>k<\/em><\/sup> are p<\/em>, 2p<\/em>, 3p<\/em>, …, p<\/em>k<\/em> \u2212 1<\/sup>p<\/em> = p<\/em>k<\/em><\/sup>, and there are p<\/em>k<\/em> \u2212 1<\/sup> of them. Therefore, the other p<\/em>k<\/em><\/sup> \u2212 p<\/em>k<\/em> \u2212 1<\/sup> numbers are all relatively prime to p<\/em>k<\/em><\/sup>.<\/p>\n\n\n\n

    Proof of Euler’s product formula<\/strong><\/p>\n\n\n\n

    The fundamental theorem of arithmetic<\/a> states that if n<\/em> > 1 there is a unique expression for n<\/em>, \"{\\displaystyle where p<\/em>1<\/sub> < p<\/em>2<\/sub> < … < p<\/em>r<\/em><\/sub> are prime numbers and each k<\/em>i<\/em><\/sub> \u2265 1. (The case n<\/em> = 1 corresponds to the empty product.)<\/p>\n\n\n\n

    Repeatedly using the multiplicative property of \u03c6<\/em> and the formula for \u03c6<\/em>(p<\/em>k<\/em><\/sup>) gives<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    This is Euler’s product formula.<\/p>\n\n\n\n

    The importance of GENERATORS OF FINITE CYCLIC GROUP<\/strong> lies in the fact that if one of the generators of a cyclic group is known, then it gets relatively easier to find the other generators of that group. We illustrate this with the help of the example of \ud835\udc48(50) under multiplication modulo 50.<\/p>\n\n\n\n

    Now, \ud835\udc48(50) = {1,3,7,9,11,13,17,19,21,23, 27,29,31,33,37,39,41,43,47,49} and therefore |\ud835\udc48(50)| = 20. Further, from the table it is easy to see that \ud835\udc48(50) is a cyclic group with 3 as one of its generators. Since 1, 3, 7, 9, 11, 13, 17, 19 are relatively prime to 20(=|\ud835\udc48(50)|), therefore we have 31, 33, 37, 39, 311, 313, 317, 319 that are all the generators of \ud835\udc48(50).<\/p>\n\n\n\n

    \ud835\udc48(50) = \u23293\u232a = \u232927\u232a = \u232937\u232a = \u232933\u232a = \u232947\u232a = \u232923\u232a = \u232913\u232a = \u232917\u232a.<\/strong><\/p>\n\n\n\n

    \"GENERATORS<\/figure>\n\n\n\n

    Now the question to be answered is how many generators an infinite cyclic group would have and what are they.<\/em><\/p>\n\n\n\n

    Theorem.<\/strong> Order of every non-identity element in an infinite cyclic group is infinite.<\/p>\n\n\n\n

    GENERATORS OF INFINITE CYCLIC GROUP<\/strong><\/p>\n\n\n\n

    Let\ud835\udc3a = \u2329\ud835\udc4e\u232a be a cyclic group of infinite order. Then \ud835\udc3a has precisely two generators \ud835\udc4e and \ud835\udc4e\u22121.<\/p>\n\n\n\n

    Proof.<\/strong> Since \ud835\udc4e\ud835\udc4e is a generator, therefore \ud835\udc4e\u22121 is also a generator of \ud835\udc3a. Thus it is enough to prove that no element other than \ud835\udc4e and \ud835\udc4e\u22121 is a generator of \ud835\udc3a. Let \ud835\udc4f \u2208 \ud835\udc3a be any generator of \ud835\udc3a. Then \ud835\udc3a = \u2329\ud835\udc4e\u232a = \u2329\ud835\udc4f\u232a and therefore there exist \ud835\udc5d, \ud835\udc5e \u2208 \u2124 such that \ud835\udc4e = \ud835\udc4f^\ud835\udc5dand
    \ud835\udc4f = \ud835\udc4e^\ud835\udc5e . Consider<\/p>\n\n\n\n

    \ud835\udc4e = \ud835\udc4f^\ud835\udc5d = (\ud835\udc4e^\ud835\udc5e )^\ud835\udc5d = \ud835\udc4e^\ud835\udc5d
    \u21d2 \ud835\udc4e^\ud835\udc5d \u22121 = \ud835\udc52
    \u21d2 \ud835\udc5d \u2212 1 = 0 [Since|\ud835\udc4e|infinite]
    \u21d2 \ud835\udc5d = \ud835\udc5e = 1 or \ud835\udc5d = \ud835\udc5e = \u22121.
    Thus either \ud835\udc4f = \ud835\udc4e or \ud835\udc4f = \ud835\udc4e\u22121 and hence \ud835\udc4e and \ud835\udc4e\u22121 are precisely the generators of \ud835\udc3a.<\/p>\n\n\n\n

    We can summarize the above results as follows- <\/em><\/p>\n\n\n\n

    Let \ud835\udc3a be a group and let \ud835\udc4e be any element of \ud835\udc3a. Then,<\/p>\n\n\n\n

    1. \u2329\ud835\udc4e\u232a = \u2329\ud835\udc4e\u22121\u232aand |\ud835\udc4e| = |\ud835\udc4e\u22121| = |\u2329\ud835\udc4e\u232a|.<\/li>
    2. |\ud835\udc4e| = \ud835\udc5b \u21d4 \u2329\ud835\udc4e\u232a = {\ud835\udc52, \ud835\udc4e, \ud835\udc4e^2, \u2026 , \ud835\udc4e^\ud835\udc5b\u22121}<\/li>
    3. \ud835\udc4e^\ud835\udc58 = \ud835\udc52 \u21d4 |\ud835\udc4e|divide \ud835\udc4e^\ud835\udc58<\/li>
    4. \ud835\udc3a is finite cyclic group \u27f9 |\ud835\udc4e| divides |\ud835\udc3a|<\/li>
    5. |\ud835\udc3a| = \ud835\udc5d (prime) and \ud835\udc4e(\u2260 \ud835\udc52) \u2208 \ud835\udc3a \u27f9 \ud835\udc3a = \u2329\ud835\udc4e\u232a<\/li>
    6. \ud835\udc3a = \u2329\ud835\udc4e\u232a finite cyclic group. Then \ud835\udc3a = \u2329\ud835\udc4e\ud835\udc58 \u232a \u21d4 \ud835\udc54.\ud835\udc50.\ud835\udc51(\ud835\udc58, \ud835\udc5b) = 1.<\/li><\/ol>\n\n\n\n\t
      \n\t\t\n\t\t\t

      \n\t\t\t\t\n\t\t\t<\/p>\n\t\t\t\t\t\t\t\t\t\n