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It is given that every element of a group generates a cyclic subgroup. I am not able to see how. If, let's say, $H=\langle a \rangle, a\in G$ then, $H={\{...,-a^{-2},-a^{-1},a^0,a^1,a^2,...\}}$. Then how is $H$ necessarily a subset of $G$ ? And further, a subgroup ?

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3 Answers 3

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Since $G$ is a group, for every $a\in G$ and $n \in \mathbb Z$ we have $a^n \in G$ (closure of the group operation). So $H=<a>$ is indeed a subset of $G$. It is a subgroup, since $a^0=e_G \in H$, and for all $a^n,a^k \in H$ we have $a^na^k=a^{n+k}\in H$ by definition. The same goes for the inverse. If $a^n\in H$, then its inverse $a^{-n}$ is also in H.

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  • $\begingroup$ got it ! Thanks $\endgroup$ Commented Sep 8, 2013 at 8:22
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$H$ will have to be a subset otherwise your group $G$ does not satisfy the closure condition.

You can check that $H$ is a subgroup by checking the necessary conditions (3 of them):

  • identity which is $e = a^0\in H$
  • inverse $a^n$ has inverse $a^{-n}\in H$ (check that $a^n\cdot a^{-n} = a^{-n}\cdot a^n = e$)
  • closure as $a^n\cdot a^m = a^{n+m} \in H$ as $n+m \in \mathbb N$

The proofs are straightforward. You can conclude that $H$ = $<a>$ is a cyclic subgroup of $G$.

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Let a belongs to group G, where G is defined by binary operation *. Then a * a belongs to G, similarly (a * a ) * a belongs to G, extending this to a times k belongs to G and thus from these elements subgroup H can be generated which satisfies closure and other properties of group.

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