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If I find the elements generated by $8$, can I say that the set of these elements is a subgroup of $\mathbf{Z}_9$ provided all these elements are in $\mathbf{Z}_9$? or to show that the set of these elements is a subgroup of $\mathbf{Z}_9$, should I test it for being a subgroup? if your answer is that no test is needed, then why is this so?

Edit: $\mathbf{Z}_9=\{0,1,2,3,4,5,6,7,8\}$ (the group under addition modulo $9$).

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    What is $z_9$ supposed to be? Integers modulo $9$?2011-04-02
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    Your first question doesn't make much sense: "the elements generated by $1$" must mean "in whatever group I am working on", and by *definition* "the elements generated by $1$" means "the elements in the smallest subgroup that contains $1$". That means it *is* a subgroup, **by definition**. Unless yb "generated by $1$" you mean something else entirely.2011-04-02
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    @Arturo: I have seen "the elements generated by $x$ as $\{x^n | n\in \mathbb{Z}\}$ (using multiplicative notation for the group operation).2011-04-02
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    Sorry for the bad form of the question. I edited the question.2011-04-02
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    @Vafa: it depends on your definition of "elements generated by $8$". Under some definitions, it is a subgroup by definition; under others, it is not. However, I think you will find that it is painfully obvious that the "elements generated by $8$" in $\mathbb{Z}_9$ form a subgroup.2011-04-02
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    @Vafa: Perhaps you should verify that "the set of elements generated by $x$ (in $G$)" is **always** a subset of $G$ and **always** a subgroup of $G$, for any group $G$ and any element $x$.2011-04-02

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The set of elements generated by $1$ is necessarily a subgroup. In fact, if $\mathbb{Z}_9$ refers to the group of integers mod $9$ with addition, this subgroup is the entire group.

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    Edited the question. What about with the edited question?2011-04-02
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    @Vafa: Did you even try? The set of elements generated by $8$ is *also* all of $\mathbb{Z}_9$.2011-04-02
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    simply because $8=-1$ in ${\Bbb Z}_9$.2011-04-02
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How many times do you need to add eight to itself to get a multiple of nine? In other words, $9|x8 \Rightarrow ?|x$. The answer will be the order of the subgroup generated by 8.