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

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Let G be the group of bespeak 5.

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



With Lagrange"s Theorem, you have the right to easily display that any kind of group of element order $p$ must it is in cyclic. I.e., any group of prime order has actually NO proper, non-trivial subgroups, due to the fact that there is no hopeful integer divisor the a prime $p$ other than $1 \text and\; p$.

That would apply to teams of order $5$.

It follows that any kind of group of stimulate $5$ (and any kind of group of prime order) have to be created by a single element and is hence, cyclic.

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


Hint because that the second problem: permit $G$ have actually order $6$. We have actually some facet $a$ of order $3$ and some facet $b$ of stimulate $2$ by Cauchy"s theorem. Present that $e,ab,(ab)^2,(ab)^3,(ab)^4,(ab)^5$ space all distinct, therefore $ab$ generates $G$.


Hint: walk you know that because that finite groups, the order of a subgroup always divides the stimulate of the group?


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