# Thread: Cyclic Groups

1. ## Cyclic Groups

Hello,
If A={1,a,a^2,a^3,a^5} is a set defined under a binary operation * can this be considered as a cyclic group.
Since a^4 is not in the set A,I have concluded that,a*(a^3)=a^4 is not in the set A. And therefore * is not a binary operation,hence this is not a group.
Is this conclution correct.Therefore in my mind a cyclic group must contain all the powers of the generator(without missing powers inbetween) i.e: The set A should be {1,a,a^2,a^3,a^4,a^5} in order to call a cyclic group under *.
Are these things I have assumed correct?
Please answer my question as quick as you can.

Thank you.

2. Going with the usual definition then yes you're missing a^4, but notice that I could define a cyclic group $\displaystyle G= \{ 1,a^2,a^4,a^6\}$ where the exponents are taken modulo $\displaystyle 8$ then clearly $\displaystyle G$ is a group although I'm missing some exponents (what I could do to get rid of this is take an isomorphism with $\displaystyle C_4$)

3. Originally Posted by Sudharaka
Hello,
If A={1,a,a^2,a^3,a^5} is a set defined under a binary operation * can this be considered as a cyclic group.
Since a^4 is not in the set A,I have concluded that,a*(a^3)=a^4 is not in the set A. And therefore * is not a binary operation,hence this is not a group.
Is this conclution correct.Therefore in my mind a cyclic group must contain all the powers of the generator(without missing powers inbetween) i.e: The set A should be {1,a,a^2,a^3,a^4,a^5} in order to call a cyclic group under *.
Are these things I have assumed correct?
Please answer my question as quick as you can.

Thank you.
yes you can show this by contradiction, suppose A is a cyclic group, then by definitions of group, if a,b are members of group G, then so is a*b, however a*a^3 does not belong to A.

4. Originally Posted by Jose27
Going with the usual definition then yes you're missing a^4, but notice that I could define a cyclic group $\displaystyle G= \{ 1,a^2,a^4,a^6\}$ where the exponents are taken modulo $\displaystyle 8$ then clearly $\displaystyle G$ is a group although I'm missing some exponents (what I could do to get rid of this is take an isomorphism with $\displaystyle C_4$)
Dear Jose27,

Can you please tell me what do you mean by taking an isomorphism. Since I have taken Group theory recently I hope you will explain it in a simple manner.

Thank you.

5. Originally Posted by Sudharaka
Hello,
If A={1,a,a^2,a^3,a^5} is a set defined under a binary operation * can this be considered as a cyclic group.
Since a^4 is not in the set A,I have concluded that,a*(a^3)=a^4 is not in the set A. And therefore * is not a binary operation,hence this is not a group.
Is this conclution correct.Therefore in my mind a cyclic group must contain all the powers of the generator(without missing powers inbetween) i.e: The set A should be {1,a,a^2,a^3,a^4,a^5} in order to call a cyclic group under *.
Are these things I have assumed correct?
Please answer my question as quick as you can.

Thank you.
Surely defining * to be $\displaystyle <a: a^3=1>$ will give you a cyclic group over this set. This will essentially say that $\displaystyle a^5 = a^2$, so we can just ignore it...

I believe the answer is meant to be "no", but this set will form a group under this operation. There is nothing that I know of in the axioms that says every element in the set must be distinct with respect to the operation.

I think.

6. Originally Posted by Sudharaka
Dear Jose27,

Can you please tell me what do you mean by taking an isomorphism. Since I have taken Group theory recently I hope you will explain it in a simple manner.

Thank you.
An isomorphism is a formal way of saying two group are "the same". For instance, taking the set $\displaystyle G_1 = \{1, 2, 3, 4, 5, 6\}$ under multiplication modulo 7 will give us a group. It is cyclic, of order 6. However, this is "the same" as the group given by the set $\displaystyle G_2 = \{0, 1, 2, 3, 4, 5\}$ under addition modulo 6, as both are cyclic of order 6. We say these groups are isomorphic. Formally, we map $\displaystyle \theta: G_1 \rightarrow G_2$, $\displaystyle 3 \mapsto 1$, $\displaystyle 3^a \text{ mod } 7 = a \text{ mod} 6$.

Now, because this mapping preserves size (it is a bijection - the image and pre-image have the same cardinality) we are half way there. We also need it to preserve the operation. That is, we require $\displaystyle (g\theta)(h\theta) = (gh)\theta$. We call such mappings homomorphisms. This holds for our mapping $\displaystyle \theta$.

A bijective homomorphism is called an isomorphism. It preserves both size and the operation.

See here for wiki's definition. You will, undoubtedly, cover this very soon in your course.

7. Dear Jose27,sym0110 and Swlabr,

Thank you so mush for the time you had spent answering these questions. Now I have a clear idea about cyclic groups.Thanks a lot.