If G has no proper subgroups, prove that G is cyclic of order p, where p is a prime number.
Assume that G has no proper subgroups, i.e., if H is a subgroup of G, then H={e} or H=G.
If G={e}, then {e} is clearly cyclic. (e)={e}.
If G=/= {e}, then ther eis an a in G with a =/= e. Then (a) is a subgroup of G, (a) =/= {e}. Then our assumption that G has no proper subgroups implies that (a)=G, so G is cyclic. Actually, the cyclic subgroup of any element of G is equal to G (since there can be no proper subgroups).
Then (a) is a subgroup of G, where (a)=/={e}. Then our assumption that G has no proper subgroups, implies that (a)=G, so G is cyclic.
For contradiction, assume that such a group is of nonprime order. So G={e,a,a^2,...,a^(n1)} where n=qb with q,b in Z, and where q and b are not equal to 1 or n (since n is not prime, this must be possible). Since b < qb1, a^b must be an element of G. According to the assumption, (a^b)=G.
(a^b) = {e, a, a^2, ... , a^(n1)}. So there must be an element k in G such that kb= qb1 (mod n) (this is of course a congruence sign, not an equal sign).
So n divides qb1kb. So there is an integer c such that
qbqkb= nc
qb1kb=(qb)c
b(qk)1=c(qb)
We know that b(qk)1 < b(qk) < qb < c(qb). So b(qk)1 < c(qb).
So the equality sign is not true, and this is a contradiciton.
Is my proof correct? If not, can you please tell me why?
Thanks in advance
Can anybody check my proof?
Announcements  Posted on  

Why bother with a post grad course  waste of time?  17102016 

 Follow
 1
 12052012 22:31

 Follow
 2
 12052012 22:40
Hi, it gets a bit complicated towards the end and I'd have just shown that a^b has order q by construction.
In fact, I kind of don't understand it, could you explain further? 
 Follow
 3
 12052012 22:50
I don't like the bit about justifying that the order of G is prime  you can just say something like 
Let g be a generator of G. Suppose that G is not prime; so G = ab for some a,b both at least 2. Then <g^a> is a cyclic subgroup of G with order b.
(If you think it's too much to just write down that g^a has order b so <g^a> is cyclic of order b, it's enough to show the weaker statement that g^a has order more than one and at most b, so the order of <g^a> is strictly between 2 and b)Last edited by matt2k8; 12052012 at 22:58. 
 Follow
 4
 12052012 23:17
(Original post by Jodin)
Hi, it gets a bit complicated towards the end and I'd have just shown that a^b has order q by construction.
In fact, I kind of don't understand it, could you explain further?
At first, I show that H is either equal to {e} or to to G.
When H=G, all the cyclic subgroups of the elements of G equal to G (because there are no proper subgroups of G). The question says that we have to show that such a group has order p, where p is prime.
I tried to use contradiction. My proof tried to show that such a group (where H=G) can exist without have order p (which is a contradiction of course).
So G = {e, a, a^2, ... ,a^(n1)}. Since n is nonprime, there exist two integers q and b (where q=/=1 or n and b=/=1 or n), such that n=qb.
Now I have to prove that all cyclic groups in G are equal to G (since G has no proper subgroups). I chose to look at a^b (we know that a^b is in G, since b < n1 = qb1) and show that it cannot equal to G, since there is no integer k, such that kb=qb1 (mod n).
This is how I tried to show it:
kb=qb1 (mod n) ==> n divides qb1kb ==> there exists an integer "c", such that nc=qb1kb. So...
nc=qb1kb
(qb)c= b(qk)1
We know that b(qk)  1 < b(qk) < qb < (qb)c. So b(qk)1 < (qb)c. In other words, this is an inequality. So my assumption that there exists a "c" such that nc=qb1kb is wrong. Which means that there is no k such that kb=qb1 (mod n). Which, in turn, means that the cyclic subgroup of a^b does not contain the last element of G, which is a^(n1). So this must be a proper subgroup...but the type of group that is described in the question says there are no proper subgroups. This is a contradiction...
Do you think my proof is correct? If not, can you tell me where I'm going wrong? Thanks. 
 Follow
 5
 12052012 23:19
(Original post by Artus)
This is what I was trying to say:
At first, I show that H is either equal to {e} or to to G.
When H=G, all the cyclic subgroups of the elements of G equal to G (because there are no proper subgroups of G). The question says that we have to show that such a group has order p, where p is prime.
I tried to use contradiction. My proof tried to show that such a group (where H=G) can exist without have order p (which is a contradiction of course).
So G = {e, a, a^2, ... ,a^(n1)}. Since n is nonprime, there exist two integers q and b (where q=/=1 or n and b=/=1 or n), such that n=qb.
Now I have to prove that all cyclic groups in G are equal to G (since G has no proper subgroups). I chose to look at a^b (we know that a^b is in G, since b < n1 = qb1) and show that it cannot equal to G, since there is no integer k, such that kb=qb1 (mod n).
This is how I tried to show it:
kb=qb1 (mod n) ==> n divides qb1kb ==> there exists an integer "c", such that nc=qb1kb. So...
nc=qb1kb
(qb)c= b(qk)1
We know that b(qk)  1 < b(qk) < qb < (qb)c. So b(qk)1 < (qb)c. In other words, this is an inequality. So my assumption that there exists a "c" such that nc=qb1kb is wrong. Which means that there is no k such that kb=qb1 (mod n). Which, in turn, means that the cyclic subgroup of a^b does not contain the last element of G, which is a^(n1). So this must be a proper subgroup...but the type of group that is described in the question says there are no proper subgroups. This is a contradiction...
Do you think my proof is correct? If not, can you tell me where I'm going wrong? Thanks. 
 Follow
 6
 12052012 23:20
Also you need to exclude the case where G has infinite order.

 Follow
 7
 12052012 23:24
(Original post by matt2k8)
I don't like the bit about justifying that the order of G is prime  you can just say something like 
Let g be a generator of G. Suppose that G is not prime; so G = ab for some a,b both at least 2. Then <g^a> is a cyclic subgroup of G with order b.
(If you think it's too much to just write down that g^a has order b so <g^a> is cyclic of order b, it's enough to show the weaker statement that g^a has order more than one and at most b, so the order of <g^a> is strictly between 2 and b)
Thanks in advance. 
 Follow
 8
 12052012 23:25
(Original post by Artus)
I could not understand how the order of <g^a> has order b...also how does this show that the order of G is prime?
Thanks in advance. 
 Follow
 9
 12052012 23:32
(Original post by matt2k8)
I'm showing that if the order of G is finite but not prime, we can construct a proper subgroup of G. 
 Follow
 10
 13052012 00:22
(Original post by matt2k8)
I don't like the bit about justifying that the order of G is prime  you can just say something like 
Let g be a generator of G. Suppose that G is not prime; so G = ab for some a,b both at least 2. Then <g^a> is a cyclic subgroup of G with order b.
(If you think it's too much to just write down that g^a has order b so <g^a> is cyclic of order b, it's enough to show the weaker statement that g^a has order more than one and at most b, so the order of <g^a> is strictly between 2 and b)
For example, if we have a cyclic group of order 6, then the order of <g^2> is 3...because the order of g^2 is x, where g^2x = g^6 (since g^6 is equal to the identity). Is that right?
Also, why do we not consider the possibility of G being infinite? 
 Follow
 11
 13052012 08:52
(Original post by Artus)
Actually, I think I undertand why...
For example, if we have a cyclic group of order 6, then the order of <g^2> is 3...because the order of g^2 is x, where g^2x = g^6 (since g^6 is equal to the identity). Is that right?
Also, why do we not consider the possibility of G being infinite?
We do need to consider the possibility of G being infinite.Last edited by matt2k8; 13052012 at 09:02. 
 Follow
 12
 13052012 12:34
Please learn how to title threads. You asked exactly the same question recently yet it is hard to find because all of your threads are entitled "Can you check this?", "Can you give me a hint?" or something equally pointless and nonspecific. This is a forum where every thread is asking for help or hints on something... narrow it down a bit  and don't just put "abstract algebra" either. Be as specific as possible so that you only need duplicate your thread title when you duplicate the thread as in this example.
Try naming the thread, for example, as "A group with no proper subgroups is cyclic of prime order" etc.

Here are the mistakes, which I am almost sure are the same as pointed out last time:
i) This is only true if is nontrivial; the trivial group is cyclic but not of prime order.
ii) You need to prove (rather than assume) that the group is of finite order i.e.
If has infinite order, then is isomorphic to the integers. But then the cyclic subgroup is a nontrivial proper subgroup for each integer equivalent to the subgroups under the obvious isomorphism . So, has finite order.
iii) "Since b < qb1, a^b must be an element of G."
Nonsense. Any power of any element of any group is an element of the group. This is by definition of a binary operation (or the closure axiom if you prefer).
iv) "So there must be an element k in G such that kb= qb1 (mod n) (this is of course a congruence sign, not an equal sign)"
You mean, an element .
v) "We know that b(qk)1 < b(qk) < qb < c(qb). So b(qk)1 < c(qb)".
You don't know whether k or c are positive or negative. 
 Follow
 13
 14052012 02:25
(Original post by Jake22)
Please learn how to title threads. You asked exactly the same question recently yet it is hard to find because all of your threads are entitled "Can you check this?", "Can you give me a hint?" or something equally pointless and nonspecific. This is a forum where every thread is asking for help or hints on something... narrow it down a bit  and don't just put "abstract algebra" either. Be as specific as possible so that you only need duplicate your thread title when you duplicate the thread as in this example.
Try naming the thread, for example, as "A group with no proper subgroups is cyclic of prime order" etc.

Here are the mistakes, which I am almost sure are the same as pointed out last time:
i) This is only true if is nontrivial; the trivial group is cyclic but not of prime order.
ii) You need to prove (rather than assume) that the group is of finite order i.e.
If has infinite order, then is isomorphic to the integers. But then the cyclic subgroup is a nontrivial proper subgroup for each integer equivalent to the subgroups under the obvious isomorphism . So, has finite order.
iii) "Since b < qb1, a^b must be an element of G."
Nonsense. Any power of any element of any group is an element of the group. This is by definition of a binary operation (or the closure axiom if you prefer).
iv) "So there must be an element k in G such that kb= qb1 (mod n) (this is of course a congruence sign, not an equal sign)"
You mean, an element .
v) "We know that b(qk)1 < b(qk) < qb < c(qb). So b(qk)1 < c(qb)".
You don't know whether k or c are positive or negative.Last edited by Artus; 14052012 at 02:29. 
 Follow
 14
 14052012 03:07
(Original post by Jake22)
Please learn how to title threads. You asked exactly the same question recently yet it is hard to find because all of your threads are entitled "Can you check this?", "Can you give me a hint?" or something equally pointless and nonspecific. This is a forum where every thread is asking for help or hints on something... narrow it down a bit  and don't just put "abstract algebra" either. Be as specific as possible so that you only need duplicate your thread title when you duplicate the thread as in this example.
Try naming the thread, for example, as "A group with no proper subgroups is cyclic of prime order" etc.

Here are the mistakes, which I am almost sure are the same as pointed out last time:
i) This is only true if is nontrivial; the trivial group is cyclic but not of prime order.
ii) You need to prove (rather than assume) that the group is of finite order i.e.
If has infinite order, then is isomorphic to the integers. But then the cyclic subgroup is a nontrivial proper subgroup for each integer equivalent to the subgroups under the obvious isomorphism . So, has finite order.
iii) "Since b < qb1, a^b must be an element of G."
Nonsense. Any power of any element of any group is an element of the group. This is by definition of a binary operation (or the closure axiom if you prefer).
iv) "So there must be an element k in G such that kb= qb1 (mod n) (this is of course a congruence sign, not an equal sign)"
You mean, an element .
v) "We know that b(qk)1 < b(qk) < qb < c(qb). So b(qk)1 < c(qb)".
You don't know whether k or c are positive or negative.
So, we know that qb1=kb (mod n). So:
qb1kb=nc
qb1kb=qbc
qbkbqbc=1
If we divide the whole equation by "b",
qkqc=1/b
Since we know that q, k, and c are integers, qkqc must also be an integer. However 1/b is a fraction, since we know that b is not equal to 1.
Do you think it is correct now?
Thanks in advance. 
 Follow
 15
 14052012 13:44
(Original post by Artus)
I think I found another way to prove it...
So, we know that qb1=kb (mod n). So:
qb1kb=nc
qb1kb=qbc
qbkbqbc=1
If we divide the whole equation by "b",
qkqc=1/b
Since we know that q, k, and c are integers, qkqc must also be an integer. However 1/b is a fraction, since we know that b is not equal to 1.
Do you think it is correct now?
Thanks in advance.
Now that you have worked out how to do it  I would also try to write it out in a more streamlined fashion. Remember, at university level (or even much before that) an answer shouldn't just be right or wrong  one should strive for a good answer. You will also learn much more if you try and write answers well.Last edited by Jake22; 14052012 at 13:48.
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: May 14, 2012
Share this discussion:
Tweet
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.