A cyclic group G is a group that can be generated by a single element. In particular, if $ G = \{ a, b, c, d, .. \} $, $ * $ is the group operation and $ a $ is a generating element, then if we compute $a $ , $a*a$, , $a*a*a $, etc. we will be able to create all members of the set G.
Get motivated - Problem from TIFR Entrance
Suppose G is a cyclic group with 60 elements. How many generators are there?
Which one of the following is TRUE?(A) \(\Bbb Z_n\)is cyclic if and only if nis prime (B) Every proper subgroup of \(\Bbb Z_n\)is cyclic (C) Every proper subgroup of \(S_4\)is cyclic (D) If every proper subgroup of a group is cyclic, then the group is cyclic.
Start with hints
Hint 1:
We will solve this question by the method of elimination. Observe that if n is prime then \(\mathbb{Z}_n\) is obviously cyclic as any of the subgroup <a> has order either 1 or n by Lagrange's theorem. Now if the order is 1 then a=id. So choose a(\(\neq\)e) \(\in \mathbb{Z}_n\) then |<a>|=n and <a> \(\subseteq\) \(\mathbb{Z}_n\) \(\Rightarrow\) <a>= \(\mathbb{Z}_n\). The problem will occur with the converse see \(\mathbb{Z}_6\) is cyclic but 6 is not prime. In general \(\mathbb{Z}_n\) = <\(\overline{1}\)> is always cyclic no matter what n is!! so option (A) is false. Can you rule out option (C)
Hint 2:
Consider option (C) every proper subgroup of \(S_4\) is cyclic.
Consider { e , (12)(34) , (13)(24) , (14)(23) } = G
Observe that this is a subgroup and |G|=4. Moreover o(g)=2 \(\forall\) g(\(\neq\)e) \(\in\) G
So G is not cyclic. Hence option (C) is not correct. Can you rule out option (D)?
Hint 3:
Consider \(\mathbb{Z}_2\)*\(\mathbb{Z}_2\) which is also known as Klein's 4 group then it is not cyclic but all of it's proper subgroups are {0}*\(\mathbb{Z}_2\) , \(\mathbb{Z}_2\)*{0} and {0}*{0} which are cyclic. Hence we can rule out option (D) as well.
Hint 4:
So option (B) is correct. Now let prove that H \(\leq\) \(\mathbb{Z}_n\) = {\(\overline{0}\),\(\overline{1}\),.....,\(\overline{n-1}\)}. By well ordering principle H has a minimal non zero element 'm'. Claim: H=<m> clearly <m> \(\subset\) H. For any r \(\in\) H by Euclid's algorithm we have r=km+d where 0 \(\leq\) d < m
which \(\Rightarrow\) d=r-km \(\in\) H
If d \(\neq\) 0 then d<m which is a contradiction
So, d=0 \(\Rightarrow\) r=km \(\Rightarrow\) H=<m> and we are done
Connected Program at Cheenta
The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.
Similar Problems
Multiplicative group from fields: TIFR GS 2018 Part A Problem 17
Understand the problem
The multiplicative group \(F^*_7\) is isomorphic to a subgroup of the multiplicative group \(F^*_{31}\).
Start with hints
Hint 1
We will write them as (Z/7Z)* and (Z/31Z)* respectively instead of the notations used.
Observe that (Z/7Z)* has order 6 and (Z/31Z)* has order 30.So there is a possibility that (Z/7Z)* is a subgroup of (Z/31Z)* by Lagrange’s Theorem.
So we need to go into the structure of the groups to solve this problem.Hence we proceed!
Let us investigate the group (Z/7Z)*.It consists of {1,2,3,4,5,6 mod 7}.Observe that 3 mod 7 generates the group.
So naturally the next question is that whether (Z/31Z)* rather is there any general result?
Hint 2
In fact the following theorem is true and describes the cyclicity (Z/nZ)* to some extent.
Theorem: If p is a prime then (Z/pZ)* is cyclic. (Check!) {Check the bonus question for the complete characterization of cyclicity of (Z/nZ)* done by Gauss.}
Hint 3
So (Z/7Z)* and (Z/31Z)* are cyclic groups of order 6 and 30 respectively with generators say A and B respectively.
Now take the element \(B^5\).The following Lemma describes its order.
Lemma: If g is the generator of the cyclic group of order n. Then \(g^k\_ has order n/gcd(n,k).(Check !)
So \(B^5\) has order 6 and hence it is isomorphic to (Z/7Z)*.
Hence the answer is True.
Hint 4
Bonus Problem:
Theorem: The group (Z/nZ)* is cyclic if and only if n is $1, 2, 4,p^k$ or 2.$p^k$, where p is an odd prime and k > 0. This was first proved by Gauss. (Wow!)
Solve and Salvage if Possible.
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Connected Program at Cheenta
The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.
Similar Problems
TIFR 2013 problem 5 | Non-Cyclic Subgroup of \(\mathbb{R}\)
Try this problem from TIFR 2013 problem 5 based on Non-Cyclic Subgroup of \(\mathbb{R}\).
Question: TIFR 2013 problem 5
True/False?
All non-trivial proper subgroups of \((\mathbb{R},+)\) are cyclic.
Hint: What subgroups comes to our mind immediately?
Discussion: \((\mathbb{Q},+)\) is a subgroup of \((\mathbb{R},+)\). Is \((\mathbb{Q},+)\) a cyclic group?
Suppose \((\mathbb{Q},+)\) is cyclic. Then there exists a generator say \(\frac{a}{b}\). Note that, we are only allowed to use addition (and subtraction) to create \(\mathbb{Q})\
Therefore, we can create $$ \frac{a}{b}+\frac{a}{b}+...+\frac{a}{b}=n\frac{a}{b}=\frac{na}{b} $$
Also, we can create $$ (-\frac{a}{b})+(-\frac{a}{b})+...+(-\frac{a}{b})=n(-\frac{a}{b})=-\frac{na}{b} $$
Notice that we can increase the magnitude of the numerator, but not the magnitude of the denominator.
For example, we cannot create \(\frac{a}{2b}\) using \(\frac{a}{b}\) and the binary operation +.
Therefore, \((\mathbb{Q},+)\) is not cyclic.
Remark: There is one result which states that subgroups of \((\mathbb{R},+)\) are either cyclic or dense. Notice that although \((\mathbb{Q},+)\) is not cyclic it is dense.
There is an element of order 51 in the multiplicative group (Z/103Z)
True
Discussion: First note that (Z/103Z) has 102 elements as 103 is a prime (in fact one of the twin primes of 101, 103 pair). Also 102 = 2317. So it has Sylow-3 subgroup of order 3 (prime order hence it is cyclic too) and a Sylow-17 subgroup (which is similarly cyclic). Since (Z/103Z) is abelian all it's subgroups are normal. Thus product of Sylow-3 and Sylow-17 subgroups is a subgroup (direct product of normal subgroups is a subgroup) containing 51 elements which is again cyclic. Hence there is an element of order 51 (generator of this subgroup).
Non trivial Proper subgroups of additive group of real numbers
All non-trivial proper subgroups of (R, +) are cyclic.
False
Discussion: There is a simple counter example: (Q, +) (the additive group of rational numbers). We also note that every additive subgroup of integers is cyclic (in fact they are of the for nZ).
Cyclic groups have exactly one generator. We can construct numerous counter examples by constructing subgroups having more than one generator. Hence another counter example is $latex a + b \sqrt 2 $ which has two generators.
Why does this does not work for integers? For example can we say that {ax + by | a, b are given integers and x, y are arbitrary} form a subgroup with two generators? No because by Bezout's Theorem we know that ax + by is merely the multiples (positive and negative) of the g.c.d(a,b).