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

You are watching: How to prove a group is cyclic

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?

Thanks because that contributing an answer to juniorg8.comematics stack Exchange!

But avoid

Asking for help, clarification, or responding to various other answers.Making statements based on opinion; earlier them increase with referrals or an individual experience.

Use juniorg8.comJax to format equations. Juniorg8.comJax reference.

See more: Clue Answer Pre Easter Period Crossword Clue, Clue Answer Pre

let $G$ be a team of bespeak $pq$, with $p$ and $q$ prime. Prove the the stimulate of the center of $G$ is 1 or $pq$.
A non-cyclic limited abelian group includes a subgroup isomorphic to the straight product the a group of bespeak $p$ with itself