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Prove that a group of order 5 must be cyclic, and every Abelian group of order 6 will also be cyclic.

Let G be the group of order 5.

To prove group of order 5 is cyclic do we have prove it by every element $(\langle a\rangle =\langle e,a,a^2,a^3,a^4,a^5=e\rangle)\forall a \in G$

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Use Lagrange's theorem. – copper.hat Mar 25 '13 at 15:46
@Serkan Cauchy's theorem will definitely be overkill for a problem like this. – Tobias Kildetoft Mar 25 '13 at 15:47
up vote 9 down vote accepted

With Lagrange's Theorem, you can easily show that any group of prime order $p$ must be cyclic. I.e., any group of prime order has NO proper, non-trivial subgroups, since there is no positive integer divisor of a prime $p$ other than $1 \text{ and}\; p$.

That would apply to groups of order $5$.

It follows that any group of order $5$ (and any group of prime order) must be generated by a single element and is hence, cyclic.

N.B. Anytime you can show that a group is generated by one element: i.e. that there exists a $g \in G$ such that $G = \langle g \rangle$, then you have proven (indeed by definition) that $G$ is cyclic.

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Nice Amy! :^))) – Babak S. Mar 25 '13 at 18:17

Hint for the second problem: Let $G$ have order $6$. We have some element $a$ of order $3$ and some element $b$ of order $2$ by Cauchy's theorem. Show that $e,ab,(ab)^2,(ab)^3,(ab)^4,(ab)^5$ are all distinct, so $ab$ generates $G$.

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to prove group is cyclic is it enough to show that one element generates the whole group – TLE Mar 25 '13 at 15:59
How to prove these all elements are different – TLE Apr 2 '13 at 15:30

Hint: Did you know that for finite groups, the order of a subgroup always divides the order of the group?

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let G contains e,a,b,c,d now every element generate a group of size 5 except e.because we cannot generate subgroup of size 2,3,4. Is my solution is correct – TLE Mar 25 '13 at 16:17
You only need to find one element that generates a group of size 5. – Jim Mar 25 '13 at 16:36

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